Solutions: Class 12 Chemistry Chapter 1 Hinglish mein

Chemistry Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

115 min
1 simulation
58 practice questions
30 MCQs

Is chapter mein aap kya seekhenge

Binary solutions ko solute aur solvent ki physical states ke hisaab se classify karo aur har type ka ek example do
Solution ki concentration ko mass, volume ya mass-by-volume percentage, ppm, mole fraction, molarity aur molality mein express karo, ek se doosre mein convert karo, aur batao kaun se temperature ke saath badalte hain
Samjho ki solute aur solvent ka nature, temperature aur pressure solubility par kaise asar daalte hain, aur gases ke liye Henry's law use karo
Do volatile liquids ke mixtures aur non-volatile solutes ke solutions par Raoult's law apply karo, vapour ki composition samet
Ideal aur non-ideal solutions mein fark karo, positive aur negative deviations ko intermolecular forces se jodo, aur azeotropes ko describe karo
Chaaron colligative properties calculate karo aur unse solute ka molar mass nikaalo
Osmosis, osmotic pressure, isotonic, hypertonic aur hypotonic solutions, aur reverse osmosis ko samjhao
Dissociation aur association ko correct karne ke liye van't Hoff factor use karo, aur abnormal molar masses aur degree of dissociation ya association nikaalo

NCERT Chemistry Class 12, Chapter 1, pp. 1–30 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

Oral rehydration salts ka packet kehta hai ki use theek ek litre saaf paani mein khaali karo. Aadha litre lo to drink bahut concentrated ho jaata hai: beemar bachche ko rehydrate karne ki jagah woh gut se paani bahar kheench sakta hai. Jewellers us gold par "22K" stamp karte hain jo mass ke hisaab se lagbhag 91.6 % gold hota hai, baaki copper aur silver. Poore India mein petrol pumps ab E20 bechte hain, yaani petrol jismein volume ke hisaab se 20 % ethanol hai. Har case mein mixture ka behaviour is baat par depend karta hai ki usmein har ingredient theek kitna hai.

Yeh unit aise hi mixtures ke baare mein hai, khaaskar liquid wale. Pehle hum solutions ko types mein baantenge aur seekhenge ki chemists unki concentration kin alag-alag tareekon se batate hain. Phir hum poochhenge ki kitna solid ya gas dissolve hoga, liquid mixture ke upar vapour ke saath kya hota hai, aur kyun kuch mixtures 'ideally' behave karte hain aur kuch nahi. Aakhir mein hum colligative properties se milenge, jo sirf is baat par depend karti hain ki kitne solute particles maujood hain, aur unse itne chhote molecules ka weight bhi nikaalenge jo dikhte nahi.

1.1 Types of Solutions

Paani mein ek chammach cheeni ghol do aur ek minute baad har ghoont barabar meetha lagta hai. Ek solution (विलयन) bilkul yahi hai: ek homogeneous mixture, jiski composition aur properties usmein har jagah same hoti hain, molecules ke scale tak. Matmaila paani yeh test fail kar deta hai: silt neeche baith jaati hai aur glass ka bottom top se alag ho jaata hai.

Solution ke parts ke naam hum unke role se rakhte hain. Solvent aam taur par woh component hota hai jo sabse zyada amount mein ho, aur wahi poore solution ki physical state tay karta hai: namak wala paani liquid hai kyunki paani liquid hai. Baaki har component ek solute hai. Is unit mein hum lagbhag poori tarah binary solutions se deal karenge, jinmein sirf do components hote hain: ek solvent aur ek solute.

Definition

Solution

Do ya zyada components ka homogeneous mixture; jo component uski physical state tay karta hai (aam taur par sabse zyada amount wala) woh solvent hai, aur baaki solutes hain.

ExampleBand kamre ki hawa, cheeni ka syrup, aur steel ke chammach ka alloy, sab solutions hain.

Dono components mein se har ek gaseous, liquid ya solid ho sakta hai, isliye binary solution ke nau possible kinds bante hain. Solute chahe jo bhi ho, solvent ki state tay karti hai ki hum solution ko gaseous, liquid ya solid kahenge (Fig. 1.1), aur Table 1.1 har kind ka ek example deta hai.

4/4Teen states, ek idea: mixture particles ke scale par uniform hai.

Fig. 1.1Har solution mein solute particles solvent particles ke beech barabar faile hote hain; solvent tay karta hai ki yeh gas mein hota hai, liquid mein ya solid mein.NCERT p. 2

Teen panels solutions ko particles ke scale par dikhate hain; circles solvent particles hain aur squares solute particles. Step 1 mein gaseous solution highlight hai: ek band jar mein door-door kuch particles, label humid air. Step 2 mein liquid solution highlight hai: beaker mein particles bheed mein lekin bikhre hue, label salt water. Step 3 mein solid solution highlight hai: particles ek regular grid mein, aur grid ki kuch jagahein solute particles ne li hain, label steel. Step 4 mein teeno panels saath dikhte hain; har ek mein solute particles solvent particles ke beech barabar faile hain.

Table 1.1Binary solution ke nau kinds, solvent ki state ke hisaab se groupedNCERT Table 1.1 · p. 2

Solution	Solute	Solvent	Example
Gaseous	gas	gas	chamakti sign tube ke neon mein mila argon
Gaseous	liquid	gas	monsoon ke din humid air mein water vapour
Gaseous	solid	gas	band almirah ki hawa mein naphthalene (mothball) ka vapour
Liquid	gas	liquid	soda water mein carbon dioxide
Liquid	liquid	liquid	paani mein ethanoic (acetic) acid, vinegar ke roop mein
Liquid	solid	liquid	nimbu-pani mein cheeni aur namak
Solid	gas	solid	metal-alloy canister ke andar store ki gayi hydrogen (ek metal hydride)
Solid	liquid	solid	silver–tin alloy mein mercury, dental amalgam ke roop mein
Solid	solid	solid	iron mein chromium aur nickel, stainless steel ke roop mein

1.2 Expressing Concentration of Solutions

Hum chai ke cup ko 'halki' ya 'strong' keh sakte hain, aur syrup ko 'patla' ya 'gaadha', lekin aise words ka matlab alag logon ke liye alag hota hai. Drip taiyaar karne wale pharmacist ya fuel blend karne wale engineer ko ek number chahiye. Solution ki concentration batati hai ki solvent ya solution ki di gayi amount mein kitna solute hai, aur chemists iske kai measures use karte hain. Har ek alag kaam ke liye theek hai, isliye sab seekho aur unke beech convert karne ki practice karo.

Fig. 1.2 ek hi solution, $100 g$ paani mein ghule $6.0 g$ urea, ko neeche diye measures se hokar follow karta hai.

6/6 $60^{\circ} C$ par liquid expand hota hai: molarity girti hai, molality nahi.

Fig. 1.2Ek solution, uski concentration batane ke paanch tareeke. Sirf volume par bana measure hi solution garam karne par badalta hai.NCERT p. 2

Left mein ek beaker mein solution hai; right mein ek card har step mein uski concentration ka ek measure dikhata hai. Step 1 recipe dikhata hai: 6.0 g urea aur 100 g paani milkar 106 g solution dete hain. Step 2 mass percentage dikhata hai, 6.0 ko 106 se divide karke 100 se multiply, lagbhag 5.7 percent. Step 3 moles mein amounts dikhata hai, 0.10 mol urea aur 5.55 mol paani, jisse urea ka mole fraction 0.018 aata hai. Step 4 molality dikhata hai: 0.10 mol per 0.100 kg paani, yaani 1.0 mol per kg. Step 5 molarity dikhata hai: solution ka volume 104 mL hai, isliye 0.96 mol per litre. Step 6 mein beaker 60 degrees Celsius tak garam kiya gaya hai: liquid level thoda upar uthta hai, molarity girkar 0.94 mol per litre ho jaati hai, jabki molality 1.0 mol per kg hi rehti hai.

Percentages aur parts per million

(i) Mass percentage (w/w). Yeh sirf masses compare karta hai, isliye balance se naapna aasaan hai aur industry mein aur product labels par common hai. $12 g$ namak aur $88 g$ paani se bana nimbu ke achaar ka brine mass ke hisaab se 12 % namak hai.

mass % = \frac{m_{A}}{m_{solution}} \times 100

(1.1)

jahan

$m_{A}$: solution mein component A ka mass
$m_{solution}$: solution ka total mass

(ii) Volume percentage (V/V). Liquids mila kar bane solutions ke liye aksar volumes naapna aasaan hota hai. '70 % isopropyl alcohol (v/v)' label wale hand sanitiser mein tayyar liquid ke har $100 m L$ mein $70 m L$ alcohol hota hai. Dhyaan do ki divisor solution ka volume hai: mix karne par liquids ke volumes hamesha theek-theek add nahi hote.

volume % = \frac{V_{A}}{V_{solution}} \times 100

(1.2)

jahan

$V_{A}$: component A ka volume
$V_{solution}$: solution ka total volume

(iii) Mass by volume percentage (w/V). Hospitals aur pharmacies solution ke per $100 m L$ solute ke grams pasand karte hain. '5 % dextrose' drip mein fluid ke har $100 m L$ mein $5 g$ glucose hota hai.

(iv) Parts per million (ppm). Traces ke liye percentages mein zeros ki lambi awkward strings aati hain, isliye hum parts per million parts ginte hain. Percentages ki tarah, ppm bhi mass/mass, volume/volume ya mass/volume ho sakta hai. Aaj ki hawa mein volume ke hisaab se carbon dioxide lagbhag 420 ppm hai, aur $1 k g$ paani mein $1 m g$ pesticide mass ke hisaab se 1 ppm hai. Water-quality aur air-pollution limits aam taur par isi tarah likhi jaati hain, ya $m g L^{- 1}$ mein, jo dilute paani ke samples ke liye lagbhag ppm jaisa hi hai.

ppm = \frac{parts of A}{all parts} \times 10^{6}

(1.3)

Mole fraction, molarity aur molality

(v) Mole fraction. Chemistry molecule-dar-molecule hoti hai, isliye solutions ke kai laws moles mein likhe jaate hain. Kisi component ka mole fraction, jise $x$ se likhte hain aur component ko subscript mein, maujood saare moles mein uska hissa hai.

x_{A} = \frac{mol of A}{total mol}

(1.4)

A aur B ke binary solution ke liye jismein $n_{A}$ aur $n_{B}$ moles hain, aur kitne bhi components wale solution ke liye:

x_{A} = \frac{n_{A}}{n_{A} + n_{B}}

(1.5)

\begin{array}{c} x_{i} = \frac{n_{i}}{n_{1} + n_{2} + \dots + n_{i}} \\ = \frac{n_{i}}{\sum n_{i}} \end{array}

(1.6)

Kyunki denominator mein har mole ek baar gina jaata hai, saare components ke mole fractions ka total theek 1 hota hai. Binary solution mein agar tumhe $x_{A}$ pata hai to $x_{B} = 1 - x_{A}$ bhi pata hai. Vapour pressures aur gas mixtures ke liye mole fractions hi natural language hain, jaisa Sections 1.4 aur 1.5 dikhayenge.

x_{1} + x_{2} + \dots + x_{i} = 1

(1.7)

(vi) Molarity (M). Kisi solution ki molarity ek litre (ek cubic decimetre) solution mein solute ke moles ki sankhya hai. Lab mein '0.10 M HCl' wali bottle ke har litre mein $0.10 m o l$ hydrogen chloride hota hai. Molarity convenient hai kyunki pipettes aur burettes se volumes naapna aasaan hai.

molarity = \frac{n_{solute}}{V_{solution}}

(1.8)

jahan

$V_{solution}$: solution ka volume litres mein

(vii) Molality (m). Molality solvent ke per kilogram (solution ke nahi) solute ke moles ki sankhya hai. $0.50 m o l k g^{- 1}$ (0.50 m) sugar solution $1 k g$ paani mein $0.50 m o l$ sugar ghol kar banta hai, volume chahe jitna bhi bane.

molality = \frac{n_{solute}}{w_{solvent}}

(1.9)

jahan

$w_{solvent}$: solvent ka mass kilograms mein

Temperature badalne par kaun sa measure bharose layak hai? Garam karne par masses aur mole numbers nahi badalte, lekin volumes badalte hain, kyunki liquids expand hote hain. Isliye mass percentage, mass ke hisaab se ppm, mole fraction aur molality fixed rehte hain, jabki solution garam hone par molarity (aur koi bhi volume-based measure) thoda gir jaata hai. Isi wajah se Section 1.6 ke laws, jo temperatures ki ek range par use hote hain, molality mein likhe jaate hain.

Compare

Solution garam karne par kya measure badalta hai?

Temperature se independent

mass percentage (w/w)
mass ke hisaab se ppm
mole fraction
molality

Temperature ke saath badalta hai

molarity
volume percentage
mass by volume percentage
per litre solution wala koi bhi measure

Example 1.1NCERT Example 1.1 · p. 4

Mass percentage se mole fraction

Car ke cooling liquid mein paani mein mass ke hisaab se 40 % propylene glycol ( $C_{3} H_{8} O_{2}$ ) hai. Har component ka mole fraction nikaalo. $M (C_{3} H_{8} O_{2}) = 76 g m o l^{- 1}$ aur $M (H_{2} O) = 18 g m o l^{- 1}$ use karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 1.2NCERT Example 1.2 · p. 4

Mass aur volume se molarity

Ek student $2.65 g$ anhydrous sodium carbonate ( $N a_{2} C O_{3}$ , $M = 106 g m o l^{- 1}$ ) paani mein gholta hai aur volumetric flask mein volume $200 m L$ tak poora karta hai. Molarity kitni hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 1.3NCERT Example 1.3 · p. 5

Non-aqueous solvent mein molality

$40.0 g$ toluene mein $1.28 g$ naphthalene ( $C_{10} H_{8}$ ) ghol kar bane solution ki molality kitni hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Density ki madad se convert karna

Sodium chloride ( $M = 58.5 g m o l^{- 1}$ ) ke 12.0 % (w/w) aqueous solution ki density $1.086 g m L^{- 1}$ hai. Iski molality aur molarity nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

Ek solution par 0.50 m label hai aur doosre par 0.50 M. Kaun sa statement sahi hai?

0.50 m label ek kilogram solvent ki baat karta hai; 0.50 M label ek litre solution kiDono mein per kilogram paani 0.50 mol solute haiDono same hain, kyunki ek litre paani ka weight ek kilogram hota hai0.50 M label ek kilogram solution ki baat karta hai

Answer dekho

Answer (A) 0.50 m label ek kilogram solvent ki baat karta hai; 0.50 M label ek litre solution ki

Chhota m molality hai (per kg solvent); capital M molarity hai (per litre solution). Yeh sirf dilute aqueous solutions ke liye paas-paas hote hain, jahan ek litre solution mein lagbhag ek kilogram paani hota hai.

Practice · 5 questions

Practice: concentration ke sawaal (Intext Questions 1.1–1.5): answers ke saath

NumericalEasyNCERT Intext Question 1.1
$132 g$ hexane mein $18.0 g$ toluene ghola gaya hai. Har component ka mass percentage nikaalo.
Solution dikhao
AnswerToluene 12.0 %, hexane 88.0 %.
Total mass $= 18.0 + 132 = 150 g$ . Toluene: $18.0 / 150 \times 100 = 12.0 %$ . Hexane: $132 / 150 \times 100 = 88.0 %$ (ya $100 - 12.0$ ).
NumericalMediumNCERT Intext Question 1.2
Chloroform ( $C H C l_{3}$ , $119.5 g m o l^{- 1}$ ) mein acetone ( $C_{3} H_{6} O$ , $58 g m o l^{- 1}$ ) ka ek solution mass ke hisaab se 25.0 % acetone hai. Acetone ka mole fraction nikaalo.
Solution dikhao
$100 g$ mein: acetone $25.0 / 58 = 0.431 m o l$ ; chloroform $75.0 / 119.5 = 0.628 m o l$ . Isliye $x_{acetone} = 0.431 / 1.059 = 0.407$ aur $x_{chloroform} = 0.593$ .
3
NumericalMediumNCERT Intext Question 1.3
Har solution ki molarity nikaalo.
1. $25.0 g$ blue vitriol, $C u S O_{4} \cdot 5 H_{2} O$ ( $249.7 g m o l^{- 1}$ ), ghol kar $2.00 L$ solution banaya gaya.
  M
  Solution dikhao
  $n = 25.0 / 249.7 = 0.1001 m o l$ , isliye molarity $= 0.1001 / 2.00 = 0.0501 M$ . Water of crystallisation bas solvent mein mil jaata hai.
2. $1.5 M$ nitric acid ke $20.0 m L$ ko paani se $250 m L$ tak dilute kiya gaya.
  M
  Solution dikhao
  Acid ke moles $= 1.5 \times 0.0200 = 0.030 m o l$ ; nayi molarity $= 0.030 / 0.250 = 0.12 M$ . (Yahi baat $M_{1} V_{1} = M_{2} V_{2}$ se bhi.)
NumericalMediumNCERT Intext Question 1.4
$0.200 m o l k g^{- 1}$ aqueous glucose solution ka $1.50 k g$ banane ke liye kitna glucose ( $180 g m o l^{- 1}$ ) chahiye?
g
Solution dikhao
$1 k g$ paani ko $0.200 m o l = 36.0 g$ glucose chahiye, jisse $1036 g$ solution banta hai. $1500 g$ solution tak scale karo: glucose $= 36.0 \times 1500 / 1036 = 52.1 g$ ( $1448 g$ paani ke saath).
5
NumericalHardNCERT Intext Question 1.5
Sodium nitrate ( $N a N O_{3}$ , $85.0 g m o l^{- 1}$ ) ka ek aqueous solution mass ke hisaab se 15.0 % hai aur uski density $1.107 g m L^{- 1}$ hai. Uski concentration ke teen measures nikaalo.
1. Molality
  mol/kg
  Solution dikhao
  $0.0850 k g$ paani mein $n = 15.0 / 85.0 = 0.1765 m o l$ : $m = 2.08 m o l k g^{- 1}$ .
2. Molarity
  M
  Solution dikhao
  $V = 100 / 1.107 = 90.3 m L$ , isliye molarity $= 0.1765 / 0.0903 = 1.95 M$ .
3. $N a N O_{3}$ ka mole fraction
  Solution dikhao
  Paani: $85.0 / 18 = 4.722 m o l$ . $x = 0.1765 / (0.1765 + 4.722) = 0.0360$ .

1.3 Solubility

Thande paani ke glass mein sugar daalte raho aur ek point aata hai jab chahe jitna bhi hilao, crystals bina ghule neeche pade rehte hain. Kisi substance ki solubility uski woh sabse badi amount hai jo diye gaye temperature par (aur gases ke liye, pressure par) solvent ki batayi gayi amount mein ghul sakti hai. Yeh is par depend karti hai ki solute aur solvent kya hain, aur temperature aur pressure par. Pehle hum liquids mein solids dekhenge, phir gases.

Definition

Solubility

Kisi substance ki sabse badi quantity jo solvent ki di gayi quantity batayi gayi temperature aur pressure par solution mein rakh sakti hai.

ExampleRoom temperature par $100 g$ paani mein lagbhag $36 g$ common salt ghulta hai.

1.3.1 Solubility of a Solid in a Liquid

Haldi ghee ke tadke ko itna gehra rang kyun deti hai, lekin paani ko bas halka sa? Uska yellow pigment, curcumin, zyadatar non-polar molecule hai, aur woh paani ke mukaable non-polar fat mein kahin behtar milta hai. Common salt ulta karta hai: woh paani mein aasaani se ghul jaata hai aur coconut oil mein bilkul nahi. Petrol aur kerosene grease ke daag utha dete hain jinhe paani chhoo bhi nahi paata. Thumb rule hai 'like dissolves like': solute us solvent mein achchhe se ghulta hai jiske intermolecular forces uske apne jaise hon, polar mein polar aur non-polar mein non-polar.

Dekho ki molecular level par kya hota hai jab ek solid solvent mein pada hota hai. Particles crystal surface chhod kar liquid mein phailte hain; yeh dissolution hai. Usi samay, crystal ke paas ghoomte ghule hue particles wapas us par chipak jaate hain; yeh crystallisation hai. Shuru mein dissolution jeetta hai, lekin jaise-jaise solution bharta hai, crystallisation tez hota jaata hai, jab tak dono rates barabar nahi ho jaate (Fig. 1.3). Uske baad ghuli hui amount constant rehti hai halaanki dono processes chalte rehte hain: yeh dynamic equilibrium hai.

Dissolution equilibrium

solute + solvent ⇌ solution

(1.10)

3/3Dono rates barabar hain. Ghuli hui amount ab nahi badalti, halaanki dono processes chalte rehte hain.

Fig. 1.3Dissolution aur crystallisation ek saath chalte hain; saturated solution woh hai jahan unke rates barabar ho jaate hain.NCERT p. 6

Left mein ek beaker ke bottom mein crystals ka dher hai aur ghule hue particles dots se dikhaye gaye hain; right mein do bars dissolving ke rate ko crystallising ke rate se compare karte hain. Step 1 mein dher bada hai, ghule hue particles kam hain, dissolving bar lamba hai aur crystallising bar bahut chhota. Step 2 mein dher chhota hai, ghule hue particles kahin zyada hain, aur crystallising bar badh gaya hai jabki dissolving bar waisa hi hai. Step 3 mein kuch crystals ab bhi bache hain, ghule hue particles badhna ruk gaye hain, aur dono bars ki length barabar hai: solution saturated hai.

Bina ghule solute ke saath equilibrium ki is state mein jo solution hai, woh saturated solution (संतृप्त विलयन) hai: us temperature aur pressure par usmein jitna solute aa sakta hai utna hai, isliye uski concentration hi solubility hai. Jo solution abhi aur solute ghol sakta hai woh unsaturated hai.

Definition

Saturated solution

Bina ghule solute ke saath dynamic equilibrium mein ek solution; us temperature aur pressure par usmein solute ki maximum possible amount hoti hai.

Temperature ka effect. Kyunki equation (1.10) ek equilibrium hai, Le Chatelier's principle batata hai ki garam karne par woh kaise respond karega. Agar lagbhag saturated solid ko gholne mein heat absorb hoti hai ( $Δ_{sol} H > 0$ , endothermic), to temperature badhane se equilibrium right ki taraf jaata hai aur solubility badhti hai. Agar gholne mein heat release hoti hai ( $Δ_{sol} H < 0$ , exothermic), to garam karne par solubility girti hai. Practice mein dono behaviours dikhte hain. Potassium nitrate endothermically ghulta hai aur uski solubility tezi se badhti hai, jabki common salt ke liye heat effect bahut chhota hai aur change bhi (Table 1.2).

Table 1.2Per 100 g paani grams mein approximate solubility

Temperature	Potassium nitrate	Sodium chloride
0 °C	13	35.7
20 °C	32	35.9
40 °C	64	36.4
60 °C	110	37.1
80 °C	169	38.0
100 °C	246	39.2

Pressure ka effect. Dabane se solid ya liquid ka volume lagbhag nahi badalta, isliye liquid mein kitna solid ghulta hai us par pressure ka koi noticeable effect nahi hota. Gases alag hain, jaisa hum aage dekhenge.

1.3.2 Solubility of a Gas in a Liquid

Paani mein ghulne mein gases ke beech bahut bada fark hota hai. Nitrogen mushkil se ghulti hai, aur oxygen bas thodi si, phir bhi ghuli hui oxygen ki wahi chhoti amount har machhli aur water plant ko zinda rakhti hai. Doosre extreme par ammonia itni aasaani se ghulti hai ki thande paani ka ek volume gas ke kai sau volumes le leta hai. Lekin kisi ek gas ke liye do conditions ghuli hui amount ko control karti hain: pressure aur temperature.

Pressure ka effect. Gas aur liquid ki boundary ko ek bheed wala darwaaza samjho. Gas molecules jis rate se andar jaate hain woh is par depend karta hai ki har second surface par kitne pahunchte hain, jo per unit volume gas molecules ki sankhya ke, yaani gas pressure ke, proportional hai. Jis rate se woh bahar aate hain woh is par depend karta hai ki kitne pehle se ghule hue hain. Equilibrium par dono rates barabar hote hain, isliye ghuli hui amount ko pressure ke saath-saath badhna hi padega. Gas par piston neeche dabao (Fig. 1.4) to ek pal ke liye nikalne se zyada molecules andar jaate hain; ghuli hui amount tab tak badhti hai jab tak bahar ka traffic barabar nahi ho jaata, aur system zyada pressure par zyada ghuli gas ke saath settle ho jaata hai.

3/3Aur gas ghulti hai jab tak point wapas line par nahi aata, zyada pressure aur zyada mole fraction par.

Fig. 1.4Gas ko compress karne se uska pressure badhta hai; phir aur gas ghulti hai jab tak system Henry's-law line par wapas nahi aata.NCERT Fig. 1.1 · p. 7

Left mein ek band cylinder mein liquid hai, uske upar gas aur top par piston; right mein ek graph gas ke partial pressure ko liquid mein uske mole fraction ke against plot karta hai, origin se guzarti ek straight line ke saath. Step 1 mein piston upar hai, gas kam ghani hai, kam gas particles ghule hain, aur ek point kam pressure par line par hai. Step 2 mein piston neeche dabaya gaya hai: gas bheed mein hai lekin ghuli hui amount abhi nahi badli, isliye point seedha upar, line ke upar, kood jaata hai. Step 3 mein aur gas ghul gayi hai aur point zyada pressure aur bade mole fraction par wapas line par aa gaya hai.

LawHenry's law

Fixed temperature par, solution ke saath equilibrium mein kisi gas ka partial pressure solution mein us gas ke mole fraction ke directly proportional hota hai.

Henry's law

p = K_{H} x

(1.11)

jahan

$p$: solution ke upar gas ka partial pressure
$x$: solution mein gas ka mole fraction
$K_{H}$: Henry's law constant, pressure units mein (yeh gas, solvent aur temperature par depend karta hai)

Constant $K_{H}$ gas–solvent pair ki property hai. Rearrange karne par $x = p / K_{H}$ : diye gaye pressure par, bade $K_{H}$ wali gas kam ghulti hai. Table 1.3 mein helium aur nitrogen insoluble end par hain aur carbon dioxide, jo paani se thoda react karti hai, soluble end par. Oxygen ki values yeh bhi dikhati hain ki $K_{H}$ temperature ke saath badhta hai, matlab garam paani mein gases kam ghulti hain.

Table 1.3Paani mein kuch gases ke approximate Henry's law constantsNCERT Table 1.2 · p. 8

Gas	Temperature / K	K_H / kbar
Helium	298	145
Nitrogen	298	86
Hydrogen	298	71
Methane	298	41
Oxygen	288	37
Oxygen	298	44
Oxygen	308	51
Carbon dioxide	298	1.7

Example 1.4NCERT Example 1.4 · p. 8

Ek talaab mein kitni oxygen hoti hai?

$298 K$ par talaab ke sampark mein hawa $0.21 b a r$ ke partial pressure par oxygen deti hai. $K_{H} (O_{2}) = 44 k b a r$ lekar, $1.00 L$ paani mein ghuli oxygen ki amount millimoles aur milligrams mein nikaalo. Ek litre paani ko $55.5 m o l$ $H_{2} O$ maano.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Henry's law industry mein, medicine mein aur pahadon mein dikhta hai:

Fizzy drinks. Soda water aur soft drinks carbon dioxide ke high pressure par bottle kiye jaate hain taaki bahut saari gas ghul jaaye. Dhakkan kholo to drink ke upar ka pressure hawa ke CO2 jitna gir jaata hai, jo bahut kam hai, isliye extra gas bubbles ki dhaar ban kar nikal jaati hai.
Divers aur bends. Andamans ke paas samundar mein 30 m neeche ek diver lagbhag chaar guna atmospheric pressure par hawa leta hai, isliye blood aur tissues mein lagbhag chaar guna nitrogen ghulti hai. Agar diver bahut tezi se upar aaye, to nitrogen solution se nikal kar blood vessels aur joints mein bubbles ban jaati hai, ek dardnaak aur khatarnaak condition jise bends kehte hain. Gehre divers aksar aise mixtures saans mein lete hain jinmein nitrogen ke kuch hisse ki jagah helium hoti hai, jo nitrogen se bhi kam soluble hai.
High altitudes. Leh mein, lagbhag 3500 m upar, air pressure aur isliye oxygen ka partial pressure sea level ki values ke lagbhag do-tihaai hote hain. Blood mein kam oxygen ghulti hai, aur naye aane wale kamzor, saans phoolna aur uljhan mehsoos kar sakte hain, is condition ko anoxia kehte hain; visitors ko acclimatise hone ke liye ek-do din aaram karne ko kaha jaata hai.

Temperature ka effect. Paani ka pan garam karo to ubalne se bahut pehle uski deewaron par chhote bubbles chipak jaate hain: ghuli hui hawa bahar aa rahi hai. Ghula hua gas molecule aise gas molecule jaisa hai jo liquid mein condense ho gaya ho, aur condense hone par heat release hoti hai, isliye gas ka ghulna exothermic hai. Le Chatelier's principle ke hisaab se temperature badhane se equilibrium wapas free gas ki taraf shift hota hai, aur solubility girti hai. Isiliye garam talaab mein thande talaab se kam oxygen hoti hai, aur garam, shaant dopahar mein machhliyan surface ke paas jama ho jaati hain.

Try karke dekho

Garam soda, thanda soda

Tumhe chahiye

soda ki do ek jaisi sealed bottles
ek refrigerator
ek sink

Pehle guess karo

Kholne par kaun si bottle zyada jhaag degi?

Thandi waliGaram waliDono barabar, kyunki dono mein same gas hai

Ek hi soda ki do ek jaisi bina khuli bottles lo. Ek ko kuch ghante fridge mein rakho aur doosri ko garam kamre mein chhod do.
Har bottle ko sink ke upar, chehron se door point karke pakdo, aur ek ke baad ek kholo.
Compare karo ki kitna fizz uthta hai aur kitni jaldi. Phir dono ko das minute khula chhodo aur har ek ka ek ghoont chakho.

Table 1.4Tumne kya dekha?

Bottle	Kholne par fizz	10 min baad
Thandi
Garam

Tumhari entries apne aap save hoti hain.

Practice · 2 questions

Practice: Henry's law ke sawaal (Intext Questions 1.6–1.7): answers ke saath

NumericalMediumNCERT Intext Question 1.6
$273 K$ aur $1.00 b a r$ ke carbon dioxide pressure par, paani mein $C O_{2}$ ki solubility $0.0767 m o l k g^{- 1}$ hai. Henry's law constant bar aur pascals mein calculate karo. Ise Table 1.3 mein $298 K$ wali value se compare karo.
bar
Solution dikhao
$1 k g$ mein paani ke moles: $1000 / 18.02 = 55.5 m o l$ . CO2 ka mole fraction: $x = 0.0767 / (0.0767 + 55.5) = 1.38 \times 10^{- 3}$ .
$K_{H} = p / x = 1.00 b a r / 1.38 \times 10^{- 3} = 725 b a r = 7.25 \times 10^{7} P a$ . $298 K$ par constant lagbhag $1.7 k b a r$ hai, do guna se bhi zyada, isliye thanda paani do guna se zyada CO2 gholta hai: yahi ek wajah hai ki garam hone par fizzy drink zyada flat lagti hai.
NumericalMediumNCERT Intext Question 1.7
Soda ki $750 m L$ ki bottle $298 K$ par $3.0 b a r$ ke carbon dioxide pressure par seal ki jaati hai. $K_{H} (C O_{2}) = 1.66 k b a r$ lekar aur drink ko paani maan kar, kitni CO2 (grams mein) ghuli hui hai?
g
Solution dikhao
$x = p / K_{H} = 3.0 / 1660 = 1.81 \times 10^{- 3}$ . Paani ke moles: $750 / 18.02 = 41.6 m o l$ .
$n_{C O_{2}} \approx x \times n_{water} = 1.81 \times 10^{- 3} \times 41.6 = 0.075 m o l$ , aur $0.075 \times 44 = 3.3 g$ CO2.

1.4 Vapour Pressure of Liquid Solutions

Section 1.3.2 mein liquids mein ghuli gases ki baat hui. Do tarah ke liquid solutions bache hain, aur yeh section dono ko dekhta hai: do liquids ke mixtures, jaise hand sanitiser ke alcohols, jahan har component evaporate ho sakta hai, aur liquid mein solid ke solutions, jaise sugar syrup, jahan aam taur par sirf solvent evaporate hota hai. Dono cases mein sabse important naapne layak property solution ka vapour pressure hai.

Yaad karo ki vapour pressure kya hai. Ek band flask mein liquid daalo to kuch molecules uski surface se upar ki jagah mein nikal jaate hain. Woh vapour molecules bhi surface se takraate hain aur kuch wapas pakad liye jaate hain. Vapour tab tak bharta hai jab tak nikalna aur pakda jaana same rate par na hone lage. Tab vapour jo pressure lagata hai woh liquid ka vapour pressure hai. Yeh sirf liquid aur temperature par depend karta hai, is par nahi ki liquid kitna hai, aur liquid garam hone par tezi se badhta hai. Ise Fig. 1.5 mein try karo.

SimulationBand container mein liquid ke upar vapour

Paani ke particles ka ek container jismein dhakkan, ek thermometer aur 20 se 100 degrees Celsius tak ka temperature slider hai, saath mein dhakkan hatane ya lagane ka button. Dhakkan laga ho to particles liquid surface chhod kar upar ki jagah mein ghoomte hain; unmein se kuch wapas gir jaate hain. Kuch der baad liquid ke upar unki sankhya badhna ruk jaati hai. Zyada temperature par kisi bhi pal vapour mein zyada particles hote hain. Dhakkan hata ho to nikle hue particles door chale jaate hain aur liquid dheere-dheere gayab ho jaata hai.

Fig. 1.5Dhakkan wale container mein paani. Play dabao, phir temperature badlo ya dhakkan uthao.

Try karo

40 °C par dhakkan laga kar Play dabao aur paani ke upar ki jagah dekho. Kya vapour particles ki sankhya badhti rehti hai, ya settle ho jaati hai?
Temperature 80 °C tak badhao. Jab yeh phir settle ho jaaye, to kya vapour mein pehle se zyada particles hain ya kam? Isse vapour pressure ke baare mein kya pata chalta hai?
Dhakkan hata do. Ab nikalne aur pakde jaane ke beech balance kabhi kyun nahi ban sakta?

1.4.1 Vapour Pressure of Liquid-Liquid Solutions

Do solvents se bane nail-polish remover ko soonghoge to dono ki smell aayegi: dono vapour mein hain. Do volatile liquids, 1 aur 2, ke aise mixture ko container mein seal karo aur intezaar karo. Har liquid upar ki jagah mein molecules bhejta hai jab tak evaporation aur condensation balance na ho jaayein, aur phir vapour ek total pressure $p_{total}$ lagata hai jo do partial vapour pressures, $p_{1}$ aur $p_{2}$ , se bana hai. Hamara sawaal hai ki yeh liquid ki composition par, jo mole fractions $x_{1}$ aur $x_{2}$ se naapi jaati hai, kaise depend karte hain.

French chemist François-Marie Raoult ne 1880s mein bahut se mixtures ke vapour pressures naap kar iska jawab diya. Unka result, Raoult's law, kamaal ka simple hai: har component aise evaporate hota hai jaise woh pure ho, bas liquid ke molecules mein uske hisse ke hisaab se kam.

LawRaoult's law (volatile components ke liye)

Volatile liquids ke solution mein, har component ka partial vapour pressure liquid mein uske mole fraction ke directly proportional hota hai; proportionality ka constant same temperature par us component ka pure state mein vapour pressure hai.

p_{1} = p_{1}^{\circ} x_{1}

(1.12)

jahan

$p_{1}^{\circ}$: mixture ke temperature par liquid 1 ka akele ka vapour pressure

p_{2} = p_{2}^{\circ} x_{2}

(1.13)

Dalton's law of partial pressures kehta hai ki total vapour pressure partial pressures ka sum hai. Raoult's law substitute karke, aur phir $x_{1} = 1 - x_{2}$ rakh kar, total pressure sirf ek mole fraction mein milta hai:

p_{total} = p_{1} + p_{2}

(1.14)

\begin{array}{c} p_{total} = x_{1} p_{1}^{\circ} + x_{2} p_{2}^{\circ} \\ = (1 - x_{2}) p_{1}^{\circ} + x_{2} p_{2}^{\circ} \end{array}

(1.15)

p_{total} = p_{1}^{\circ} + (p_{2}^{\circ} - p_{1}^{\circ}) x_{2}

(1.16)

Equation (1.16) ek straight line hai, aur ise dhyaan se padhne par teen conclusions milte hain. Pehla, total vapour pressure ko kisi bhi ek component ke mole fraction se likha ja sakta hai. Doosra, yeh $x_{2}$ ke saath linearly badalta hai, pure liquid 1 ke $p_{1}^{\circ}$ se pure liquid 2 ke $p_{2}^{\circ}$ tak. Teesra, component 1 aur daalne se total badhega ya ghatega, yeh is par depend karta hai ki kaun sa pure liquid zyada volatile hai: agar $p_{1}^{\circ} < p_{2}^{\circ}$ , to $x_{1}$ badhne par total girta hai. Fig. 1.6 is line ko uske do parts se banata hai.

4/4Stacks ke tops $p_{1}^{\circ}$ se $p_{2}^{\circ}$ tak ek straight line par hain: equation (1.16).

Fig. 1.6Ideal mixture ke liye har partial pressure ek straight line hai, isliye unka sum, total vapour pressure, bhi straight line hai.NCERT Fig. 1.3 · p. 10

Ek graph jismein neeche component 2 ka mole fraction 0 se 1 tak hai aur upar ki taraf vapour pressure. Pure component 1 ka vapour pressure kam hai, left axis par; pure component 2 ka zyada hai, right axis par. Step 1 component 1 ke partial pressure ki dashed line banata hai, jo left mein uski pure value se right mein zero tak girti hai. Step 2 component 2 ki dashed line jodta hai, jo left mein zero se right mein uski pure value tak badhti hai. Step 3 dono partial pressures ko mole fractions 0.25, 0.5 aur 0.75 par bars ki tarah stack karta hai, ek segment doosre ke upar. Step 4 bars ke tops ko ek solid straight line se jodta hai, total vapour pressure, jo component 1 ki pure value se component 2 ki pure value tak jaati hai.

Vapour mein kya hai? Dalton's law ke hisaab se har component ka partial pressure uska vapour mein mole fraction, jise $y$ likhte hain, guna total pressure hai. Isliye $y_{i} = p_{i} / p_{total}$ :

p_{1} = y_{1} p_{total}

(1.17)

p_{2} = y_{2} p_{total}

(1.18)

Vapour ki composition

p_{i} = y_{i} p_{total}

(1.19)

Kyunki zyada volatile liquid pressure mein apne hisse se zyada contribute karta hai, vapour hamesha us liquid se, jisse woh bana, zyada volatile component mein richer hota hai. Us vapour ko condense karke phir evaporate karo, to woh aur bhi richer ho jaata hai; fractional distillation ke peeche yahi idea hai. Ise Fig. 1.7 mein explore karo: deviation slider ko beech mein rakh kar, dono pure vapour pressures set karo, composition badlo, aur $y_{2}$ ko $x_{2}$ se compare karo.

InteractiveRaoult's law explorer

Component 2 ke mole fraction ke against vapour pressure ka ek interactive graph. Sliders dono pure liquids ke vapour pressures set karte hain (component 2 ko zero set karke non-volatile banaya ja sakta hai), liquid mein component 2 ka mole fraction, aur unlike molecules ek doosre ko like molecules ke mukaable kitni mazbooti se attract karte hain. Graph dono components ki partial pressure lines, total vapour pressure aur, mixture non-ideal ho to, comparison ke liye seedhi Raoult's-law line dikhata hai. Readouts dono partial pressures, total, aur vapour mein component 2 ka mole fraction dete hain, saath mein ek message jo batata hai ki vapour kis component mein richer hai, ya mixture azeotrope par hai. Ek preset Example 1.5 ka methanol aur ethanol mixture load karta hai.

Fig. 1.7Pure vapour pressures aur liquid composition set karo; partial aur total pressures aur vapour composition padho. Deviation slider mixture ko non-ideal banata hai (Section 1.5).

Example 1.5NCERT Example 1.5 · p. 11

Total vapour pressure aur vapour composition

Methanol aur ethanol lagbhag ideal solution banate hain. $313 K$ par unke pure vapour pressures $35.4 k P a$ aur $17.8 k P a$ hain. Ek mixture $32.0 g$ methanol ( $C H_{3} O H$ , $32 g m o l^{- 1}$ ) aur $69.0 g$ ethanol ( $C_{2} H_{5} O H$ , $46 g m o l^{- 1}$ ) se banaya jaata hai. (i) $313 K$ par mixture ka vapour pressure aur (ii) vapour mein har alcohol ka mole fraction nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

1.4.2 Raoult's Law as a special case of Henry's Law

Raoult's law, $p_{i} = p_{i}^{\circ} x_{i}$ , ko Henry's law, $p = K_{H} x$ , ke neeche likho, to dono alag constants wala ek hi law lagte hain: dono mein solution ke upar kisi component ka partial pressure usmein uske mole fraction ke proportional hai. Yeh coincidence nahi hai. Ghuli hui gas bas ek aisa component hai jo working temperature par gas banne layak volatile hai, aur Henry's constant uske vapour pressure ka role nibhata hai. Isliye Raoult's law ko Henry's law ka woh khaas case maana ja sakta hai jismein $K_{H} = p_{i}^{\circ}$ .

Compare

Do laws, ek shape

Henry's law

$p = K_{H} x$
liquid mein ghuli gas ke liye use hota hai
$K_{H}$ har gas–solvent pair ke liye experiment se milta hai

Raoult's law

$p_{i} = p_{i}^{\circ} x_{i}$
solution mein volatile liquid ke liye use hota hai
constant pure liquid ka vapour pressure hai

1.4.3 Vapour Pressure of Solutions of Solids in Liquids

Roz ke kai solutions liquid mein ghule solid hote hain: achaar ke brine mein namak, syrup mein cheeni, tincture ke roop mein alcohol mein iodine, coconut oil mein camphor. Aam taur par solid non-volatile hota hai: uske particles lagbhag kabhi vapour mein nahi nikalte. Tab solution ka poora vapour pressure solvent se aata hai, aur experiments dikhate hain ki yeh same temperature par pure solvent ke vapour pressure se hamesha kam hota hai.

Liquid surface ko exits ki ek line samjho jinse molecules vapour mein ja sakte hain (Fig. 1.8). Pure solvent mein har exit use mein hai. Non-volatile solute gholo to surface ki kuch jagahein solute particles le lete hain jo kabhi nahi nikalte, isliye har second kam solvent molecules nikalte hain. Di gayi vapour density ke liye vapour molecules ab bhi purane rate se liquid mein lautte hain, isliye nikalna aur lautna chhoti vapour density par, yaani kam vapour pressure par, balance hote hain. (Gehra treatment dikhata hai ki solvent ki escaping tendency uske mole fraction ke proportional hai; surface wali picture ise imagine karne ka convenient tareeka hai.)

3/3Chhoti vapour density par capture dheeme escape se match karta hai: kam vapour pressure.

Fig. 1.8Non-volatile solute surface ka kuch hissa le leta hai, isliye kam solvent molecules nikalte hain aur vapour pressure kam hota hai.NCERT Fig. 1.4 · p. 12

Side se dikhti liquid surface ki ek magnified strip, upar vapour ke saath. Step 1 mein pure solvent: har surface site ek gol solvent molecule hai, paanch arrows molecules ko nikalte dikhate hain, aur upar vapour kaafi bheed wala hai. Step 2 mein solute particles, squares ki tarah bane, das mein se teen surface sites le lete hain; unke koi escape arrows nahi hain, aur sirf teen solvent molecules nikal rahe hain. Step 3 mein solution ke upar ka vapour pure solvent ke upar se kam molecules ke saath settle ho gaya hai, label lower vapour pressure.

Kitna kam? Girawat is par depend karti hai ki solute kitna hai, is par nahi ki woh kya hai: aadha mole glucose aur aadha mole glycerol, har ek ek kilogram paani mein ghula, uska vapour pressure lagbhag bilkul barabar ghatate hain. Raoult's law ka general form is case ko cover karta hai: kisi bhi solution mein, har volatile component apne mole fraction ke proportion mein partial vapour pressure contribute karta hai. Solvent ko component 1 aur non-volatile solute ko component 2 maano, to sirf component 1 contribute karta hai:

Non-volatile solute ka solution

p_{1} = x_{1} p_{1}^{\circ}

(1.20)

$p_{1}$ ka $x_{1}$ ke against graph zero (koi solvent nahi) se $p_{1}^{\circ}$ (pure solvent) tak ek straight line hai. Upar wale explorer mein component 2 ka vapour pressure zero set karke tum ise dekh sakte ho.

1.5 Ideal and Non-ideal Solutions

Raoult's law kuch liquid mixtures par bilkul fit hota hai aur kuch par bura. Isse liquid–liquid solutions ko do families mein baantne ka natural tareeka milta hai: jo har composition par ise follow karte hain, aur jo nahi karte.

1.5.1 Ideal Solutions

Ek ideal solution composition ki poori range mein Raoult's law follow karta hai. Iske saath do aur properties hamesha hoti hain. Pure components ko milane par na heat release hoti hai na absorb, aur mixture ka volume milaye gaye volumes ka theek sum hota hai:

\begin{array}{c} Δ_{mix} H = 0 \\ Δ_{mix} V = 0 \end{array}

(1.21)

Definition

Ideal solution

Aisa solution jo har composition par Raoult's law follow kare, $Δ_{mix} H = 0$ aur $Δ_{mix} V = 0$ ke saath.

Molecular wajah do components A aur B ke saath sabse aasaani se dikhti hai. Pure liquids mein har molecule sirf apni hi kind se juda hota hai: A–A ya B–B attractions. Mixture mein A–B attractions bhi aa jaate hain. Agar A–B attraction lagbhag utna hi strong hai jitne A–A aur B–B attractions jinki jagah woh leta hai, to molecule 'bata' nahi sakta ki uske padosi A hain ya B. Uski escaping tendency nahi badalti, isliye Raoult's law lagta hai, milane par koi energy gain ya loss nahi hoti, aur molecules pehle jaise pack hote hain. Sach mein ideal solutions kam milte hain, lekin bahut milte-julte molecules ke mixtures paas aate hain: n-pentane aur n-hexane, chlorobenzene aur bromobenzene, aur Example 1.5 ka methanol–ethanol pair.

1.5.2 Non-ideal Solutions

Ek non-ideal solution composition ki poori range mein Raoult's law follow nahi karta: uska vapour pressure equation (1.16) ke prediction se zyada ya kam hota hai. Agar zyada hai, to solution Raoult's law se positive deviation dikhata hai; agar kam, to negative deviation (Fig. 1.9). Phir se, wajah intermolecular forces ka balance hai.

3/3A–B attractions mazboot: molecules roke jaate hain aur curve line ke neeche latakta hai.

Fig. 1.9Unlike molecules ke beech kamzor attractions vapour pressure ko Raoult's line ke upar dhakelte hain; mazboot attractions use neeche kheenchte hain.NCERT Fig. 1.6 · p. 13

Component 2 ke mole fraction ke against total vapour pressure ka graph, left aur right axes par pure-component vapour pressures ke saath. Step 1 unhe jodne wali seedhi dashed Raoult's-law line dikhata hai, label ideal, ek note ke saath ki unlike attractions like attractions jitne strong hain. Step 2 line ke upar jhukta ek curve jodta hai, label positive deviation, note ke saath ki unlike molecules ek doosre ko kamzori se attract karte hain. Step 3 line ke neeche latakta ek curve jodta hai, label negative deviation, note ke saath ki unlike molecules ek doosre ko zyada mazbooti se attract karte hain. Comparison ke liye ideal line poore samay dikhti rehti hai.

Positive deviation. Yahan A–B attractions A–A aur B–B attractions se kamzor hain. Har kind ka molecule mixture mein apne pure liquid se kam mazbooti se pakda jaata hai, aasaani se nikal jaata hai, aur vapour pressure Raoult's-law value se upar chala jaata hai. Hexane ke saath mila ethanol iska saaf example hai. Pure ethanol mein molecules hydrogen bonds se jude hote hain; hexane molecules unke beech ghus kar unmein se kai bonds tod dete hain aur badle mein sirf kamzor attractions dete hain. Hydrogen bonds todne mein energy lagti hai, isliye mixing heat absorb karta hai ( $Δ_{mix} H > 0$ ), aur dheeli packing ki wajah se aam taur par volume badhta hai ( $Δ_{mix} V > 0$ ).

Negative deviation. Yahan A–B attractions like attractions se mazboot hain. Har molecule mixture mein zyada kaske pakda jaata hai, kam aasaani se nikalta hai, aur vapour pressure Raoult's-law value se neeche girta hai. Diethyl ether ke saath chloroform ek example hai: $C H C l_{3}$ ka hydrogen, jo apne teen chlorine padosiyon ki wajah se thoda positive hai, ether ke oxygen se hydrogen bond banata hai, ek aisa attraction jo kisi pure liquid mein nahi hai. In naye attractions ke banne se energy release hoti hai, isliye mixture garam hota hai ( $Δ_{mix} H < 0$ ) aur aam taur par thoda sikudta hai ( $Δ_{mix} V < 0$ ). Hydrochloric acid aur paani, jahan strong ion–water attractions bante hain, bhi isi tarah deviate karte hain.

Compare

Ideal aur non-ideal liquid mixtures

Ideal

A–B ≈ A–A, B–B
Raoult's law follow karta hai
$Δ_{mix} H = 0$ , $Δ_{mix} V = 0$
jaise n-pentane + n-hexane

Positive deviation

A–B kamzor
vapour pressure Raoult se upar
$Δ_{mix} H > 0$ , $Δ_{mix} V > 0$
jaise ethanol + hexane
minimum-boiling azeotrope bana sakta hai

Negative deviation

A–B mazboot
vapour pressure Raoult se neeche
$Δ_{mix} H < 0$ , $Δ_{mix} V < 0$
jaise chloroform + diethyl ether
maximum-boiling azeotrope bana sakta hai

Azeotropes. Jab deviation bada hota hai, to total vapour pressure curve mein ek maximum ya minimum ban jaata hai. Us khaas composition par liquid aur usse nikalne wale vapour ki composition bilkul same hoti hai, isliye ubalne se mixture nahi badalta aur liquid pure substance ki tarah ek constant temperature par ubalta hai. Aisa mixture ek azeotrope (स्थिरक्वाथी) hai, aur kitna bhi fractional distillation use aage alag nahi kar sakta.

Bada positive deviation vapour-pressure maximum deta hai, matlab mixture dono pure liquids se neeche ubalta hai: ek minimum-boiling azeotrope. Ethanol aur paani mass ke hisaab se lagbhag 95.6 % ethanol par aisa banate hain, jo lagbhag 351 K par ubalta hai, isliye fermented liquid ko distil karke kabhi pure ethanol nahi mil sakta.
Bada negative deviation vapour-pressure minimum aur ek maximum-boiling azeotrope deta hai, jo dono pure liquids se upar ubalta hai. Hydrochloric acid aur paani mass ke hisaab se lagbhag 20 % HCl par aisa banate hain, jo lagbhag 382 K (109 °C) par ubalta hai, isiliye sirf paani uda kar concentrated hydrochloric acid nahi banaya ja sakta.

Definition

Azeotrope

Fixed composition wala liquid mixture jo constant temperature par ubalta hai aur same composition ka vapour deta hai, isliye fractional distillation use alag nahi kar sakta.

Dono kinds ke deviation Fig. 1.7 mein try karo: attraction control ko 'weaker' ya 'stronger' ki taraf slide karo aur woh point dhoondo jahan vapour aur liquid ki compositions barabar ho jaati hain.

Quick check

3 mein se 1

Do liquids milane se mixture garam ho jaata hai. Raoult's law se tum kis kind ke deviation ki ummeed karoge?

Positive, kyunki mixture garam hai aur zyada evaporate hota haiKoi nahi, kyunki garam hone se vapour pressure par asar nahi padtaPositive, kyunki molecules mein zyada energy haiNegative, kyunki heat release ka matlab naye, mazboot A–B attractions hain

Answer dekho

Answer (D) Negative, kyunki heat release ka matlab naye, mazboot A–B attractions hain

Heat release (

Δ_{mix} H < 0

) dikhata hai ki bane hue A–B attractions un A–A aur B–B attractions se mazboot hain jinki jagah unhone li. Mazboot pakad matlab kam escaping tendency: negative deviation. Same temperature par compare karne se garam hone ka effect hat jaata hai.

Practice · 1 question

Practice: liquid mixtures ke sawaal (Intext Question 1.8): answers ke saath

1
NumericalMediumNCERT Intext Question 1.8
Pure liquids P aur Q ke vapour pressures $330 K$ par $32.0 k P a$ aur $84.0 k P a$ hain, aur woh ideal solution banate hain. Inke ek mixture ka is temperature par total vapour pressure $60.0 k P a$ hai. (a) liquid aur (b) vapour ki composition nikaalo.
1. Liquid mein Q ka mole fraction
  Solution dikhao
  $60.0 = 32.0 + (84.0 - 32.0) x_{Q}$ , isliye $x_{Q} = 28.0 / 52.0 = 0.538$ aur $x_{P} = 0.462$ .
2. Vapour mein Q ka mole fraction
  Solution dikhao
  $p_{Q} = 0.538 \times 84.0 = 45.2 k P a$ , isliye $y_{Q} = 45.2 / 60.0 = 0.754$ aur $y_{P} = 0.246$ . Vapour zyada volatile Q mein richer hai.

1.6 Colligative Properties and Determination of Molar Mass

Non-volatile solute solvent ka vapour pressure kam karne se zyada kuch karta hai (Section 1.4.3). Wahi change solution ko zyada temperature par ubalne aur kam temperature par jamne par majboor karta hai, aur solvent ko sahi membrane se hokar solution mein ghusne ko bhi. Isse chaar naapne layak effects milte hain:

solvent ke vapour pressure ka relative lowering,
uske boiling point ka elevation,
uske freezing point ka depression, aur
solution ka osmotic pressure.

Ab kamaal ki baat. Dilute solution mein har effect solvent particles ke mukaable solute particles ki ginti se tay hota hai, woh particles chahe jo bhi hon: same paani mein $0.1 m o l$ glucose aur $0.1 m o l$ urea same chaaron effects dete hain, halaanki glucose molecule teen guna bhaari hai. Aise effects ko colligative properties (अणुसंख्य गुणधर्म) kehte hain, Latin colligatus se, yaani 'saath bandhe hue', kyunki yeh particles ko ek collection ki tarah lekar depend karti hain. Kyunki yeh particles ginti hain, solute ke known mass ke liye inmein se ek ko naapne se pata chalta hai ki particles ka har mole kitna bhaari hai: molar mass.

Definition

Colligative property

Dilute solution ki aisi property jo solvent particles ke mukaable solute particles ki sankhya par depend karti hai, unki chemical identity par nahi.

1.6.1 Relative Lowering of Vapour Pressure

Raoult ne dikhaya ki solvent ke vapour pressure ka lowering sirf solute particles ki concentration par depend karta hai. Equation (1.20) se shuru karke, ek chhota derivation ise molar masses nikaalne ke tool mein badal deta hai.

DerivationRaoult's law se molar mass tak

Non-volatile solute (2) wale solvent (1) ke liye, Raoult's law solvent ka vapour pressure deta hai:

p_{1} = x_{1} p_{1}^{\circ}

(1.22)

Vapour pressure ka lowering pure solvent se fark hai:

\begin{array}{c} Δ p_{1} = p_{1}^{\circ} - p_{1} \\ = p_{1}^{\circ} - x_{1} p_{1}^{\circ} \\ = p_{1}^{\circ} (1 - x_{1}) \end{array}

(1.23)

Binary solution mein $1 - x_{1} = x_{2}$ , isliye lowering solute ke mole fraction ke proportional hai:

Δ p_{1} = x_{2} p_{1}^{\circ}

(1.24)

$p_{1}^{\circ}$ se divide karne par relative lowering of vapour pressure milta hai, ek pure number jo solute ke mole fraction ke barabar hai. (Kai non-volatile solutes hon to $x_{2}$ unke mole fractions ka sum ban jaata hai.)

\frac{Δ p_{1}}{p_{1}^{\circ}} = \frac{p_{1}^{\circ} - p_{1}}{p_{1}^{\circ}} = x_{2}

(1.25)

$n_{1}$ moles solvent aur $n_{2}$ moles solute ke saath $x_{2}$ likho:

\frac{p_{1}^{\circ} - p_{1}}{p_{1}^{\circ}} = \frac{n_{2}}{n_{1} + n_{2}}

(1.26)

Dilute solution mein $n_{2} ≪ n_{1}$ , isliye denominator se $n_{2}$ hataya ja sakta hai. Phir solvent aur solute ke masses $w_{1}$ , $w_{2}$ aur molar masses $M_{1}$ , $M_{2}$ ke saath, $n = w / M$ :

\frac{p_{1}^{\circ} - p_{1}}{p_{1}^{\circ}} \approx \frac{n_{2}}{n_{1}}

(1.27)

\frac{p_{1}^{\circ} - p_{1}}{p_{1}^{\circ}} = \frac{w_{2} \times M_{1}}{M_{2} \times w_{1}}

(1.28)

Equation (1.28) mein $M_{2}$ ke alawa sab kuch naapa ja sakta hai, isliye isse solute ka molar mass milta hai.

Example 1.6NCERT Example 1.6 · p. 16

Vapour-pressure measurement se molar mass

Pure cyclohexane ( $C_{6} H_{12}$ , $84 g m o l^{- 1}$ ) ka $298 K$ par vapour pressure $13.00 k P a$ hai. Jab $60.0 g$ cyclohexane mein $2.40 g$ non-volatile, non-electrolyte solid ghola jaata hai, to vapour pressure girkar $12.66 k P a$ ho jaata hai. Solid ka molar mass nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

1.6.2 Elevation of Boiling Point

Chutki bhar namak wala chawal ka paani saade paani se degree ke kuch sauvein hisse der se ubalta hai. Kyun? Liquid tab ubalta hai jab uska vapour pressure badh kar upar ki hawa ke pressure ke barabar ho jaata hai; pure paani ke liye yeh $373.15 K$ par hota hai, jahan vapour pressure $1.013 b a r$ tak pahunchta hai. Non-volatile solute gholne se har temperature par vapour pressure kam ho jaata hai, isliye $373.15 K$ par solution abhi bhi $1.013 b a r$ se kam hai aur ubalne se pehle use aur garam karna padta hai. Isliye solution hamesha apne pure solvent se upar ubalta hai, jaisa Fig. 1.10 ke pehle teen steps dikhate hain.

5/5Solution ka curve solid ke curve se kam temperature $T_{f}$ par milta hai: depression $Δ T_{f} = T_{f}^{\circ} - T_{f}$ .

Fig. 1.10Vapour-pressure curve neeche jaane se atmosphere line se uska milan point zyada temperature par chala jaata hai, aur solid ke curve se milan point kam temperature par.NCERT Fig. 1.7 · p. 17

Ek schematic graph jismein upar ki taraf vapour pressure aur right ki taraf temperature hai. Step 1 pure liquid solvent ka ek upar jaata curve aur atmospheric pressure, 1.013 bar, par ek horizontal dashed line banata hai; jahan yeh cross karte hain woh normal boiling point hai. Step 2 solution ke liye pehle curve ke theek neeche ek doosra curve jodta hai, jiska vapour pressure har temperature par kam hai. Step 3 mark karta hai ki solution curve atmospheric line tak kahan pahunchta hai, zyada temperature par: dono crossing points ke beech ka gap boiling-point elevation hai. Step 4 dhyaan low-temperature end par le jaata hai aur solid solvent ka ek zyada steep curve jodta hai; yeh normal freezing point par pure liquid curve se milta hai. Step 5 mark karta hai ki solution curve solid curve se kahan milta hai, kam temperature par: gap freezing-point depression hai.

Vapour-pressure lowering ki tarah, yeh rise bhi solute particles ki sankhya par depend karta hai, unki kind par nahi. Dilute solutions mein measurements dikhate hain ki boiling point ka rise, jise elevation of boiling point $Δ T_{b} = T_{b} - T_{b}^{\circ}$ kehte hain, solute ki molality ke proportion mein badhta hai:

Δ T_{b} \propto m

(1.29)

Boiling-point elevation

Δ T_{b} = K_{b} m

(1.30)

jahan

$K_{b}$: solvent ka boiling point elevation constant (molal elevation ya ebullioscopic constant), $K k g m o l^{- 1}$ mein
$m$: solute ki molality

$K_{b}$ woh elevation hai jo $1 m o l k g^{- 1}$ solution ideal rehne par dikhata; yeh solvent ki property hai (Table 1.5). Agar molar mass $M_{2}$ wale solute ke $w_{2}$ grams ko $w_{1}$ grams solvent mein ghola jaaye, to molality hai

m = \frac{w_{2} / M_{2}}{w_{1} / 1000} = \frac{1000 \times w_{2}}{M_{2} \times w_{1}}

(1.31)

aur ise equation (1.30) mein substitute karne par elevation milta hai, ya ulta, naape gaye elevation se molar mass:

Δ T_{b} = \frac{K_{b} \times 1000 \times w_{2}}{M_{2} \times w_{1}}

(1.32)

M_{2} = \frac{1000 \times w_{2} \times K_{b}}{Δ T_{b} \times w_{1}}

(1.33)

Example 1.7NCERT Example 1.7 · p. 17

Sugar solution ka boiling point

Ek halwai $400 g$ paani mein $68.4 g$ sucrose ( $C_{12} H_{22} O_{11}$ , $342 g m o l^{- 1}$ ) gholta hai. $1.013 b a r$ par yeh syrup kis temperature par ubalna shuru karega? Paani ke liye $K_{b} = 0.52 K k g m o l^{- 1}$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 1.8NCERT Example 1.8 · p. 18

Sulfur molecule mein kitne atoms?

$100 g$ carbon disulphide ( $K_{b} = 2.34 K k g m o l^{- 1}$ ) mein $2.56 g$ sulfur ghola jaata hai. Boiling point $0.234 K$ badhta hai. Is solution mein sulfur ka molar mass aur har molecule mein atoms ki sankhya nikaalo (S ka atomic mass = 32.06).

Pehle khud try karo. Phir solution ek-ek step karke kholo.

1.6.3 Depression of Freezing Point

Ab Fig. 1.10 ke low-temperature end (steps 4 aur 5) ko dekho. Solid solvent ka apna vapour-pressure curve hai, jo liquid ke curve se zyada steep hai. Jahan dono curves cross karte hain, wahan solid aur liquid ke vapour pressures barabar hote hain, isliye vapour ke raaste koi bhi doosre ki keemat par badh nahi sakta; yeh crossing freezing point hai. Jab dilute solution jamta hai, to alag hone wale crystals pure solvent hote hain, isliye important yeh hai ki solution ka curve solid solvent ke curve se kahan milta hai. Solution ka curve neeche hai, isliye woh solid ke curve tak sirf kam temperature par pahunchta hai: freezing point gir jaata hai.

Dilute solutions ke liye depression of freezing point, $Δ T_{f} = T_{f}^{\circ} - T_{f}$ , molality ke proportional hai:

Freezing-point depression

Δ T_{f} = K_{f} m

(1.34)

jahan

$K_{f}$: solvent ka freezing point depression constant (molal depression ya cryoscopic constant), $K k g m o l^{- 1}$ mein

Equation (1.31) se molality substitute karne par depression milta hai, aur naape gaye depression se molar mass:

Δ T_{f} = \frac{K_{f} \times w_{2} \times 1000}{M_{2} \times w_{1}}

(1.35)

M_{2} = \frac{K_{f} \times w_{2} \times 1000}{Δ T_{f} \times w_{1}}

(1.36)

Dono constants sirf solvent par depend karte hain, aur thermodynamics inhe solvent ke molar mass $M_{1}$ ( $g m o l^{- 1}$ mein), uske freezing aur boiling points $T_{f}$ aur $T_{b}$ (kelvin mein), uski enthalpies of fusion aur vaporisation, aur gas constant $R$ se deta hai:

K_{f} = \frac{R \times M_{1} \times T_{f}^{2}}{1000 \times Δ_{fus} H}

(1.37)

K_{b} = \frac{R \times M_{1} \times T_{b}^{2}}{1000 \times Δ_{vap} H}

(1.38)

Table 1.5Kuch solvents ke ebullioscopic aur cryoscopic constants (temperatures K mein, constants K kg mol⁻¹ mein)NCERT Table 1.3 · p. 19

Solvent	Ubalta hai	$K_{b}$	Jamta hai	$K_{f}$
Water	373.15	0.52	273.15	1.86
Ethanol	351.4	1.22	159.0	1.99
Benzene	353.2	2.53	278.7	5.12
Cyclohexane	353.9	2.79	279.7	20.0
Carbon disulphide	319.4	2.34	161.6	3.83
Ethanoic acid	391.0	3.07	289.8	3.90
Camphor	–	–	452	37.7

Example

1.86 kahan se aata hai?

Paani $273.15 K$ par $Δ_{fus} H = 6.01 k J m o l^{- 1}$ ke saath pighalta hai, aur $M_{1} = 18.02 g m o l^{- 1}$ . Paani ke liye $K_{f}$ calculate karne ke liye equation (1.37) use karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 1.9NCERT Example 1.9 · p. 20

0 °C se neeche paani ko liquid rakhna

Store kiye samples ko jamne se bachane ke liye paani mein glycerol ( $C_{3} H_{8} O_{3}$ , $92 g m o l^{- 1}$ ) milaya jaata hai. $400 g$ paani mein $46.0 g$ glycerol ghola jaata hai. (a) freezing point ka depression aur (b) solution ka freezing point calculate karo. $K_{f} (water) = 1.86 K k g m o l^{- 1}$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 1.10NCERT Example 1.10 · p. 20

Rast ke camphor method se molar mass

$0.150 g$ ek organic compound ko $10.0 g$ camphor ke saath pighlaya jaata hai. Mixture ka melting point pure camphor ( $K_{f} = 37.7 K k g m o l^{- 1}$ ) se $3.47 K$ neeche hai. Compound ka molar mass nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

1.6.4 Osmosis and Osmotic Pressure

Sikude hue kishmish ko raat bhar paani mein bhigo do aur subah tak woh phool jaate hain. Raita ke liye kheere ke slices par namak chhidko aur kuch minute mein woh apne hi paani ke pool mein pade hote hain. Murjhaya palak thande paani ke bowl mein phir se taaza ho jaata hai. Har case mein ek membrane hai: kishmish ki skin, plant cells ki walls. Kaafi magnification par dekho to aisi membrane ek jaali hai jiske gaps water molecule se bas thode chaude hain. Paani nikal sakta hai; bade solute particles, aur water molecules mein lipte ions, zyadatar nahi nikal paate. Aise barrier ko semipermeable membrane kehte hain. Living cell membranes aise hi behave karti hain, aur artificial membranes bhi, jaise cellophane aur water purifiers ke andar ki patli polymer films.

Maan lo ek semipermeable membrane pure paani ko ek solution se alag karti hai, jaise Fig. 1.11 mein. Water molecules dono taraf se membrane se takraate hain, lekin solution wali side par kuch 'takkar' solute particles ki hoti hai jo paar nahi ja sakte, isliye solution mein bahar nikalne se zyada paani andar jaata hai. Semipermeable membrane se hokar solvent ka yeh net flow, pure solvent (ya dilute solution) se zyada concentrated solution mein, osmosis kehlata hai. Direction yaad rakho: solvent chalta hai, hamesha zyada solute particles wali side ki taraf.

3/3Solution par theek $Π$ se dabane par koi net flow nahi hota; levels barabar rehte hain.

Fig. 1.11Solvent solution mein tab tak behta hai jab tak pressure difference use rok na de; flow rokne ke liye zaroori pressure osmotic pressure hai.NCERT Fig. 1.9 · p. 20

Ek U-shaped tube jiske mod ke bottom mein ek semipermeable membrane hai, dashed line se bani. Left arm mein pure paani hai aur right arm mein solute particles wala solution. Step 1 mein dono levels barabar hain aur arrows dikhate hain ki left se right zyada paani jaata hai, wapas kam. Step 2 mein right ka level upar utha hai aur left ka gira hai; height ka fark h se mark hai, aur net flow ruk gaya hai kyunki extra liquid peeche dhakelta hai. Step 3 mein ek piston right arm par osmotic pressure se dabata hai, dono levels barabar rakhta hai taaki koi net flow ho hi na.

Solution par dabane se flow ruk sakta hai. Solution wali side par woh sabse chhota extra pressure jo kisi bhi solvent ko andar aane se rokta hai, solution ka osmotic pressure (परासरण दाब), $Π$ , hai. Yeh bhi colligative hai: yeh solute particles ki sankhya par depend karta hai, unki kind par nahi. Dilute solutions ke liye experiments dikhate hain ki yeh molar concentration $C$ aur absolute temperature ke proportional hai, aur proportionality ka constant gas constant $R$ hai:

Definition

Osmotic pressure

Woh extra pressure jo tumhe solution par lagana padta hai taaki semipermeable membrane se hokar usmein koi solvent na aaye; yeh naapta hai ki solution kitni mazbooti se solvent ko andar kheenchta hai.

Osmotic pressure

Π = C R T

(1.39)

jahan

$C$: solute ka molar concentration ( $m o l L^{- 1}$ )
$R$: gas constant, $0.083 L b a r m o l^{- 1} K^{- 1}$
$T$: temperature kelvin mein

$V$ litres solution mein $n_{2}$ moles solute ke liye $C = n_{2} / V$ likhne aur $n_{2} = w_{2} / M_{2}$ rakhne se molar mass tak ka raasta milta hai:

Π = \frac{n_{2}}{V} R T

(1.40)

Π V = \frac{w_{2} R T}{M_{2}}

(1.41)

M_{2} = \frac{w_{2} R T}{Π V}

(1.42)

Proteins, polymers aur doosre bade molecules ke liye equation (1.42) preferred method hai. Aise substance ke kuch grams mein bahut kam moles hote hain, isliye boiling-point aur freezing-point changes naapne ke liye bahut chhote honge, lekin osmotic pressure phir bhi kuch millibar ka padhne layak hota hai. Measurement room temperature par hota hai, jo un proteins ke liye important hai jo garam karne se kharab ho jaate, aur yeh molarity use karta hai, jo volumetric flasks mein bane solutions ke liye convenient hai.

Example 1.11NCERT Example 1.11 · p. 22

Ek protein ka molar mass

$0.350 g$ purified protein ko paani mein ghol kar $50.0 m L$ solution banaya jaata hai. $298 K$ par iska osmotic pressure $2.61 \times 10^{- 3} b a r$ hai. Protein ka molar mass nikaalo ( $R = 0.0831 L b a r m o l^{- 1} K^{- 1}$ ). Phir isi solution ke freezing-point depression ka estimate lagao.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Hospital drip '0.9 % saline' se kyun banti hai, kabhi saade paani se nahi? Kyunki red blood cells khud semipermeable membrane mein lipti solution ki chhoti-chhoti theliyan hain (Fig. 1.12). Sodium chloride ka 0.9 % (mass/volume) solution, jise normal saline kehte hain, ka osmotic pressure cells ke andar ke fluid jitna hai, isliye paani barabar andar-bahar jaata hai aur cells apni shape banaye rakhte hain. Same temperature par barabar osmotic pressure wale solutions ko isotonic kehte hain. Cells ko saade paani mein daalo, jiska osmotic pressure kam hai (ek hypotonic solution), to paani tezi se andar ghusta hai jab tak woh phool kar phat na jaayein. Unhe tez brine mein daalo, jiska osmotic pressure zyada hai (ek hypertonic solution), to paani unse bahar nikalta hai, aur woh sikud jaate hain.

1/30.9 % saline mein paani barabar andar aata aur bahar jaata hai: cell apni shape rakhta hai.

Fig. 1.12Paani cell membrane se hokar zyada osmotic pressure wali side ki taraf jaata hai.NCERT p. 22

Beech mein bada bana ek red blood cell, jiske chaaron taraf woh solution hai jismein use rakha gaya hai. Step 1 mein solution isotonic hai, 0.9 percent namak: cell apni normal dimpled disc shape rakhta hai aur chhote arrows paani ko barabar andar-bahar jaate dikhate hain. Step 2 mein solution hypotonic hai, saada paani: arrows andar point karte hain aur cell phool kar phatne ke kareeb ek gol ball ban gaya hai. Step 3 mein solution hypertonic hai, tez namak wala paani: arrows bahar point karte hain aur cell sikud kar chhoti jhurriyon wali shape ban gaya hai.

Osmosis tumhare chaaron taraf kaam kar raha hai. Paani plant roots mein kuch had tak osmosis se jaata hai, aur cells ke andar jo pressure woh banata hai woh pattiyon aur stems ko kadak rakhta hai; isiliye murjhaya palak paani mein theek ho jaata hai. Bahut zyada namak wali diet body fluids ka osmotic pressure badha deti hai, isliye body paani rok leti hai aur tissues phool sakte hain, is sujan ko edema kehte hain. Achaar, namak lagi machhli aur fruit preserves jaise aam ka murabba mahino tak tikte hain kyunki un par girne wala koi bhi bacterium ya mould ek bahut hypertonic liquid se ghira hota hai, osmosis se paani kho deta hai aur badh nahi paata.

Try karke dekho

Paani aur syrup mein kishmish

Tumhe chahiye

lagbhag 30 kishmish
do glass
cheeni aur paani
ek chammach
ek plate

Pehle guess karo

Saade paani mein kishmish ka kya hoga?

Woh aur sikud jaayengiWoh phool jaayengiKuch nahi, kyunki kishmish pehle se sookhi hain

Ek glass mein nal ka paani bharo aur doosre mein gaadha sugar syrup (aadhe glass paani mein lagbhag chaar chammach cheeni, saaf hone tak hilao).
Har glass mein milte-julte size ki das kishmish gin kar daalo. Comparison ke liye das aur ek sookhi plate par rakho.
Aath se barah ghante baad kishmish bahar nikaalo, thapthapa kar sukhao aur unka size aur chhoone par feel compare karo.

Table 1.6Bhigone ke baad kishmish

Kahan rakhi	Size	Feel
Nal ka paani
Sugar syrup
Sookhi plate

Tumhari entries apne aap save hoti hain.

1.6.5 Reverse Osmosis and Water Purification

Agar solution wali side par pressure uske osmotic pressure se zyada badha diya jaaye, to flow ulti taraf chalta hai: pure solvent solution se nichod kar membrane ke paar nikal jaata hai, aur solute peeche reh jaata hai. Yeh reverse osmosis (RO) hai, aur yahi kai Indian kitchens ke RO purifier ka working principle hai, aur un bade plants ka bhi jo samundar ke paani ko peene ke paani mein badalte hain (Fig. 1.13).

3/3Ghule hue salts peeche reh jaate hain aur reject water mein, zyada concentrated hokar, nikal jaate hain.

Fig. 1.13Reverse osmosis mein pump paani ko membrane se itni tezi se dhakelta hai jitna osmosis use wapas nahi kheench sakta; salts reject stream mein peeche reh jaate hain.NCERT Fig. 1.11 · p. 23

Ghar ke water purifier ka flow diagram. Nal ka paani left se ek pump mein aata hai, jo use high pressure par ek lambe membrane module mein bhejta hai, jo ek tube ki tarah bana hai aur lambai mein ek dashed membrane se bata hai. Step 1 mein feed water andar aata hai aur pump par label hai pressure greater than the osmotic pressure. Step 2 mein water molecules membrane paar karke neeche wale channel mein jaate hain aur purified water label wale storage tank mein nikalte hain. Step 3 mein ghule hue salts, chhote squares ki tarah bane, membrane ke upar reh jaate hain aur ek alag outlet se reject water label wale drain mein nikalte hain, feed se zyada concentrated.

Ise kaam karne layak banane ke liye das-das bar ke pressures jhel sakne wali mazboot membrane chahiye thi. Shuruaati reverse-osmosis membranes cellulose acetate ki patli films thin; ab zyadatar purifiers aur plants ek aur bhi patli polyamide layer use karte hain jo ek porous backing par bani hoti hai, aur backing strain jhelti hai. Dono tarah, film water molecules ko jaane deti hai aur ghule hue ions aur zyadatar doosri impurities ko rokti hai. Kai ghar ke purifiers har ek litre saaf karne par do-teen litre reject water drain mein bhejte hain, isliye woh paani dhone ya pocha lagane ke liye jama karna faydemand hai. Shahar ke level par, Chennai apne peene ke paani ka kuch hissa Minjur aur Nemmeli ke seawater reverse-osmosis plants se leta hai.

Example

Samundar ka paani desalinate karne ke liye minimum pressure

Samundar ke paani ko $298 K$ par per litre $35 g$ sodium chloride ( $58.5 g m o l^{- 1}$ ) maano. Har formula unit solution mein do ions deta hai (Section 1.7). Reverse-osmosis plant ko kam se kam kitna pressure lagana padega, iska estimate lagao. $R = 0.0831 L b a r m o l^{- 1} K^{- 1}$ use karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 4 questions

Practice: colligative properties ke sawaal (Intext Questions 1.9–1.12): answers ke saath

NumericalMediumNCERT Intext Question 1.9
$303 K$ par pure paani ka vapour pressure $4.24 k P a$ hai. $450 g$ paani mein $30.0 g$ glucose ( $180 g m o l^{- 1}$ ) ghola jaata hai. Is solution ke upar paani ka vapour pressure aur relative lowering nikaalo.
kPa
Solution dikhao
$n_{2} = 30.0 / 180 = 0.1667 m o l$ ; $n_{1} = 450 / 18 = 25.0 m o l$ ; $x_{2} = 0.1667 / 25.17 = 0.00662$ .
Relative lowering $= x_{2} = 0.00662$ . Isliye vapour pressure $p_{1} = 4.24 \times (1 - 0.00662) = 4.21 k P a$ hai.
NumericalMediumNCERT Intext Question 1.10
Ek hill station par pure paani $96.80^{\circ} C$ par ubalta hai. Wahan use $97.00^{\circ} C$ par ubalne ke liye $250 g$ paani mein kitna glucose ( $180 g m o l^{- 1}$ ) gholna padega? $K_{b} = 0.52 K k g m o l^{- 1}$ lo.
g
Solution dikhao
$Δ T_{b} = 0.20 K$ , isliye $m = 0.20 / 0.52 = 0.385 m o l k g^{- 1}$ . $0.250 k g$ paani mein: $n = 0.0962 m o l$ , mass $= 0.0962 \times 180 = 17.3 g$ glucose.
NumericalMediumNCERT Intext Question 1.11
$60.0 g$ benzene ka freezing point $0.80 K$ kam karne ke liye usmein kitna naphthalene ( $128 g m o l^{- 1}$ ) gholna padega? $K_{f} (benzene) = 5.12 K k g m o l^{- 1}$ .
g
Solution dikhao
$m = 0.80 / 5.12 = 0.156 m o l k g^{- 1}$ ; $n = 0.156 \times 0.0600 = 9.38 \times 10^{- 3} m o l$ ; mass $= 9.38 \times 10^{- 3} \times 128 = 1.20 g$ .
NumericalMediumNCERT Intext Question 1.12
$27^{\circ} C$ par $500 m L$ paani mein molar mass $1.6 \times 10^{5} g m o l^{- 1}$ wale polymer ke $2.0 g$ ke solution ka osmotic pressure pascals mein calculate karo. $R = 8.314 J m o l^{- 1} K^{- 1}$ use karo.
Pa
Solution dikhao
$5.0 \times 10^{- 4} m^{3}$ mein $n = 2.0 / (1.6 \times 10^{5}) = 1.25 \times 10^{- 5} m o l$ , isliye $C = 0.025 m o l m^{- 3}$ .
$Π = C R T = 0.025 \times 8.314 \times 300 = 62 P a$ : bahut chhota pressure, lekin phir bhi lagbhag 6 mm paani ki height ki tarah naapa ja sakta hai.

1.7 Abnormal Molar Masses

Colligative properties particles ginti hain, aur isse dhokha ho sakta hai. Ek student $99.0 g$ paani mein $1.00 g$ sodium chloride gholta hai, lagbhag $0.62 K$ ka freezing-point depression naapta hai, use equation (1.36) mein daalta hai aur lagbhag $30 g m o l^{- 1}$ molar mass paata hai, jo 58.5 ke formula mass ka lagbhag aadha hai. Measurement mein koi galti nahi thi. Paani mein har NaCl unit ek $N a^{+}$ aur ek $C l^{-}$ ion mein toot jaati hai, isliye solution mein formula units se lagbhag do guna particles hote hain, depression lagbhag double ho jaata hai, aur usse nikala molar mass lagbhag aadha. General rule: ions mein dissociation naape gaye molar mass ko asli se chhota bana deta hai.

Molecules team bhi bana sakte hain, jiska ulta effect hota hai. Benzene, jo low dielectric constant wala non-polar solvent hai aur acid se hydrogen bond nahi bana sakta, mein ghula ethanoic acid joda bana kar dimers banata hai, har ek do hydrogen bonds se juda: $2 C H_{3} C O O H$ ⇌ $(C H_{3} C O O H)_{2}$ .

Poora dimerisation particles ki sankhya aadhi kar dega. Tab $Δ T_{b}$ aur $Δ T_{f}$ single molecules ke liye expected values ke aadhe honge, aur unse nikala molar mass $60 g m o l^{- 1}$ ki jagah lagbhag $120$ aayega. Colligative property se mila aisa molar mass jo formula mass se kam ya zyada ho, abnormal molar mass kehlata hai. Fig. 1.14 teeno cases compare karta hai.

2/3Har unit do ions mein tootti hai: 4 units 8 particles deti hain, $i \approx 2$ .

Fig. 1.14Colligative properties particles ginti hain, isliye tootne se ginti badhti hai aur joda banne se ghatti hai.NCERT p. 24

Left mein chaar formula units ghuli hain, aur unse bane particles right mein dikhaye gaye hain, van't Hoff factor ke saath. Step 1 mein paani mein chaar glucose molecules chaar particles hi rehte hain: i barabar 1. Step 2 mein paani mein sodium chloride ki chaar units aath ions mein toot jaati hain, chaar plus aur chaar minus: i lagbhag 2. Step 3 mein benzene mein chaar ethanoic acid molecules jode bana kar do dimers banate hain, har joda do dashed hydrogen bonds se juda: i lagbhag 0.5.

1880s mein Jacobus van't Hoff ne dissociation ya association ko correct karne ke liye ek correction factor $i$ diya, jise ab van't Hoff factor kehte hain. Ise teen barabar tareekon se define kiya ja sakta hai:

van't Hoff factor

\begin{array}{c} i = \frac{normal M}{abnormal M} \\ i = \frac{observed effect}{calculated effect} \\ i = \frac{mol of particles}{mol dissolved} \end{array}

Yahan abnormal molar mass experiment se mili value hai, aur 'calculated' colligative property woh hai jo solute ke na dissociate na associate hone par expected hai. Dissociation ke liye $i > 1$ ; association ke liye $i < 1$ . Dilute hone par aqueous NaCl ya KCl ka $i$ 2 ke kareeb aata hai, jabki benzene mein ethanoic acid ka $i$ 0.5 ke kareeb. $i$ shaamil karke colligative equations ban jaati hain:

Definition

van't Hoff factor

Naapi gayi colligative property aur bina badle solute ke liye calculated value ka ratio; yaani ghule hue solute ke per mole maujood particles ke moles.

van't Hoff factor ke saath colligative properties

\begin{array}{c} \frac{p_{1}^{\circ} - p_{1}}{p_{1}^{\circ}} = i \frac{n_{2}}{n_{1}} \\ Δ T_{b} = i K_{b} m \\ Δ T_{f} = i K_{f} m \\ Π = \frac{i n_{2} R T}{V} \end{array}

Table 1.7Poore dissociation ya association ke liye expected van't Hoff factorsNCERT Table 1.4 · p. 25

Solute (solvent)	Kya hota hai	Ideal $i$
Glucose, urea (water)	molecules poore rehte hain	1
$N a C l$ , $M g S O_{4}$ (water)	har formula unit se do ions	2
$C a C l_{2}$ , $K_{2} S O_{4}$ (water)	har formula unit se teen ions	3
$A l C l_{3}$ (water)	har formula unit se chaar ions	4
Ethanoic acid (benzene)	molecules ke jode (dimers)	0.5

Strong electrolytes ki naapi gayi values bahut dilute solutions ko chhod kar in ideal numbers se thodi kam aati hain. Ulte charge wale ions ek doosre ko attract karte hain, aur kuch paas-paas jode bana kar ghoomte hain jo ek particle gine jaate hain. Yeh kami zyada charge wale ions, jaise $M g^{2 +}$ aur $S O_{4}^{2 -}$ , ke liye zyada hoti hai, aur solution dilute karne par ghatti hai.

Derivationi se degree of dissociation ya association

Dissociation. Maan lo ek aise solute ka ek mole ghola gaya jo har formula unit se $n$ particles deta, aur uska ek fraction $α$ (degree of dissociation) sach mein dissociate hota hai. Tab $1 - α$ mol poore rehte hain aur $n α$ mol ions bante hain:

\begin{array}{c} i = (1 - α) + n α \\ = 1 + (n - 1) α \\ α = \frac{i - 1}{n - 1} \end{array}

Association. Agar $n$ molecules jud kar ek bada particle banate hain aur ek fraction $α$ (degree of association) aisa karta hai, to $1 - α$ mol akele rehte hain aur $α / n$ mol associated particles bante hain:

\begin{array}{c} i = (1 - α) + \frac{α}{n} \\ = 1 - (1 - \frac{1}{n}) α \\ α = \frac{1 - i}{1 - 1 / n} \end{array}

Dimerisation ( $n = 2$ ) ke liye yeh $i = 1 - α / 2$ ban jaata hai, isliye $α = 2 (1 - i)$ .

Example 1.12NCERT Example 1.12 · p. 25

Benzene mein kitna ethanoic acid dimerise hota hai?

$50.0 g$ benzene mein $1.20 g$ ethanoic acid ( $C H_{3} C O O H$ , $60 g m o l^{- 1}$ ) ghola jaata hai, aur freezing point $1.08 K$ girta hai. $K_{f} (benzene) = 5.12 K k g m o l^{- 1}$ . Observed molar mass, van't Hoff factor aur acid ka kitna percentage dimerised hai, nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 1.13NCERT Example 1.13 · p. 26

Weak acid ki degree of dissociation aur Ka

$500 g$ paani mein $0.460 g$ methanoic (formic) acid, $H C O O H$ ( $46 g m o l^{- 1}$ ), ghola jaata hai. Freezing point $0.0406 K$ kam hota hai. $K_{f} = 1.86 K k g m o l^{- 1}$ lekar van't Hoff factor, degree of dissociation aur acid dissociation constant $K_{a}$ nikaalo. Itne dilute aqueous solution mein molality aur molarity practically barabar hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

Paani mein freezing-point depression se naapa gaya ek solute ka molar mass uske formula mass ka lagbhag ek-tihaai aata hai. Solute sabse zyada sambhav hai…

C a C l_{2}

N a C l

glucosebenzene mein ethanoic acid

Answer dekho

Answer (A) $C a C l_{2}$

Normal ka ek-tihaai molar mass matlab

i \approx 3

: har formula unit se teen particles, jaise

C a^{2 +}

+ 2

C l^{-}

ke liye. NaCl lagbhag 2 deta hai, glucose 1, aur associate hone wala acid 1 se kam.

Recap

Summary aur key points

Solution ek homogeneous mixture hai; solvent uski state tay karta hai, jisse gaseous, liquid aur solid solutions bante hain (Section 1.1).
Concentration ko mass %, volume %, mass by volume %, ppm, mole fraction, molarity ya molality mein bataya ja sakta hai; sirf volume-based measures temperature ke saath badalte hain (Section 1.2).
Saturated solution ek dynamic equilibrium hai; Le Chatelier's principle batata hai ki temperature solid ki solubility ko kaise badalta hai (Section 1.3.1).
Henry's law, $p = K_{H} x$ : gas ki solubility pressure ke saath badhti hai aur temperature ke saath girti hai; bada $K_{H}$ matlab kam soluble gas (Section 1.3.2).
Raoult's law, $p_{i} = p_{i}^{\circ} x_{i}$ ; do volatile liquids ke liye total vapour pressure composition mein linear hai aur vapour zyada volatile component mein richer hota hai (Section 1.4.1).
Raoult's law, Henry's law ka $K_{H} = p^{\circ}$ wala special case hai (Section 1.4.2).
Non-volatile solute solvent ka vapour pressure $x_{1} p_{1}^{\circ}$ tak kam karta hai (Section 1.4.3).
Ideal solutions $Δ_{mix} H = 0$ aur $Δ_{mix} V = 0$ ke saath Raoult's law follow karte hain; kamzor A–B attractions positive deviations dete hain, mazboot negative (Section 1.5).
Bade deviations minimum- ya maximum-boiling azeotropes dete hain, jinhe distillation alag nahi kar sakta (Section 1.5.2).
Colligative properties solute particles ki sankhya par depend karti hain: relative lowering of vapour pressure $= x_{2}$ , $Δ T_{b} = K_{b} m$ , $Δ T_{f} = K_{f} m$ , $Π = C R T$ (Section 1.6).
Har colligative property solute ka molar mass deti hai; osmotic pressure proteins aur polymers ke liye theek hai (Section 1.6.4).
Osmotic pressure se zyada pressure reverse osmosis chalata hai, jo paani saaf karne mein use hota hai (Section 1.6.5).
van't Hoff factor $i$ dissociation ( $i > 1$ ) aur association ( $i < 1$ ) ko correct karta hai aur dono ki degree deta hai (Section 1.7).

Practice · 46 questions

Practice questions answers ke saath (NCERT “Exercises”)

1
Short answerEasyNCERT Exercises 1.1
Dono parts ka jawab do.
1. Kya cheez mixture ko suspension ki jagah solution banati hai? Kis component ko solvent kehte hain?
  Solution dikhao
  AnswerSolution molecular scale tak homogeneous hota hai; solvent woh component hai (aam taur par sabse zyada wala) jo uski physical state tay karta hai.
  Solution mein har jagah same composition aur properties hoti hain, kyunki uske components alag-alag molecules ya ions ki tarah mile hote hain; suspension mein particles itne bade hote hain ki neeche baith jaate hain. Solvent woh component hai jo solution ki state tay karta hai, aam taur par sabse zyada amount wala; baaki sab solutes hain.
2. (b)
  Har solution ko uske type (solvent mein solute) se match karo.
  Left mein koi item tap karo, phir right mein uska match.
  Answer dekho
  Answer
  tincture of iodine : liquid mein solid
  LPG vapour (butane ke saath propane) : gas mein gas
  sterling silver (silver mein copper) : solid mein solid
  gold amalgam (gold mein mercury) : solid mein liquid
  fish tank mein oxygen : liquid mein gas
  Iodine alcohol mein ghulta hai; propane aur butane dono gases hain; silver mein copper ek solid alloy hai; mercury, ek liquid, solid gold mein ghulta hai; oxygen gas paani mein ghulti hai.
Quick answerEasyNCERT Exercises 1.2
Woh example chuno jismein gas solid mein ghuli hai.
Hydride storage canister ke metal mein absorb hui hydrogenBarf ke block mein phanse hawa ke bubblesSoda water mein carbon dioxideHawa mein dhuein ke particles
Answer dekho
Answer (A) Hydride storage canister ke metal mein absorb hui hydrogen
Hydrogen molecules atoms mein toot kar metal lattice mein barabar fail jaate hain: gaseous solute wala homogeneous solid. Barf mein bubbles ek alag phase hain, soda water liquid solution hai, aur dhuaan gas mein solid kanon ka suspension hai.
3
Short answerEasyNCERT Exercises 1.3
Concentration terms.
1. (a)
  Har term ko uski definition se match karo.
  Left mein koi item tap karo, phir right mein uska match.
  Answer dekho
  Answer
  mole fraction : component ke moles ÷ total moles
  molality : solute ke moles ÷ solvent ke kg
  molarity : solute ke moles ÷ solution ke litres
  mass percentage : component ka mass ÷ solution ka mass × 100
  Yeh equations (1.4), (1.9), (1.8) aur (1.1) hain.
2. Solution garam karne par in chaar measures mein se kaun sa badalta hai, aur kyun?
  Solution dikhao
  AnswerSirf molarity, kyunki garam karne par solution ka volume badhta hai jabki masses aur moles nahi badalte.
  Molarity ke denominator mein solution ka volume hai; garam hone par liquids expand hote hain, isliye molarity thodi girti hai. Mole fraction, molality aur mass percentage mein sirf masses ya moles hain, jo temperature ke saath nahi badalte.
NumericalMediumNCERT Exercises 1.4
Lab ka concentrated sulfuric acid mass ke hisaab se 98.0 % $H_{2} S O_{4}$ hai aur uski density $1.84 g m L^{- 1}$ hai. Uski molarity calculate karo ( $M (H_{2} S O_{4}) = 98.1 g m o l^{- 1}$ ).
M
Solution dikhao
Ek litre acid ka mass $1000 \times 1.84 = 1840 g$ hai, jismein se $0.980 \times 1840 = 1803 g$ $H_{2} S O_{4}$ hai.
$1 L$ mein moles $= 1803 / 98.1 = 18.4 m o l$ , isliye acid $18.4 M$ hai.
5
NumericalMediumNCERT Exercises 1.5
Ek aqueous urea solution ( $C O (N H_{2})_{2}$ , $60.06 g m o l^{- 1}$ ) par 15.0 % w/w label hai. Uski density $1.04 g m L^{- 1}$ hai.
1. Uski molality nikaalo.
  mol/kg
  Solution dikhao
  $0.0850 k g$ paani mein $n = 15.0 / 60.06 = 0.2498 m o l$ : $m = 2.94 m o l k g^{- 1}$ .
2. Urea ka mole fraction nikaalo (paani ka 1 minus yeh hai).
  Solution dikhao
  Paani: $85.0 / 18.02 = 4.717 m o l$ . $x_{urea} = 0.2498 / (0.2498 + 4.717) = 0.0503$ ; $x_{water} = 0.950$ .
3. Uski molarity nikaalo.
  M
  Solution dikhao
  $V = 100 / 1.04 = 96.2 m L$ ; molarity $= 0.2498 / 0.0962 = 2.60 M$ .
NumericalHardNCERT Exercises 1.6
$1.50 g$ ka ek sample potassium carbonate ( $K_{2} C O_{3}$ , $138.2 g m o l^{- 1}$ ) aur potassium hydrogencarbonate ( $K H C O_{3}$ , $100.1 g m o l^{- 1}$ ) ka mixture hai jismein dono ke moles barabar hain. Isse poori tarah react karne ke liye $0.250 M$ HCl ka kitna volume chahiye?
mL
Solution dikhao
Har ek ko $n$ mol maano: $138.2 n + 100.1 n = 1.50$ , isliye $n = 6.29 \times 10^{- 3} m o l$ .
Har carbonate ion do $H^{+}$ ions leta hai (paani aur carbon dioxide banate hue), aur har hydrogencarbonate ion ek.
Zaroori HCl $= 2 n + n = 3 n = 0.0189 m o l$ . Volume $= 0.0189 / 0.250 = 0.0755 L = 75.5 m L$ .
NumericalEasyNCERT Exercises 1.7
12.0 % (mass ke hisaab se) sugar syrup ka $250 g$ 30.0 % syrup ke $150 g$ ke saath milaya jaata hai. Mixture mein sugar ka mass percentage kya hai?
%
Solution dikhao
Sugar: $400 g$ mixture mein $0.120 \times 250 + 0.300 \times 150 = 30.0 + 45.0 = 75.0 g$ , isliye $75.0 / 400 \times 100 = 18.75 %$ (lagbhag 18.8 %).
8
NumericalMediumNCERT Exercises 1.8
Ek engine coolant $152 g$ propylene glycol ( $C_{3} H_{8} O_{2}$ , $76 g m o l^{- 1}$ ) aur $250 g$ paani se banta hai. Uski density $1.035 g m L^{- 1}$ hai.
1. Molality calculate karo.
  mol/kg
  Solution dikhao
  $n = 152 / 76 = 2.00 m o l$ ; $m = 2.00 / 0.250 = 8.00 m o l k g^{- 1}$ .
2. Molarity calculate karo.
  M
  Solution dikhao
  Solution ka mass $= 402 g$ ; $V = 402 / 1.035 = 388 m L$ ; molarity $= 2.00 / 0.388 = 5.15 M$ .
9
NumericalMediumNCERT Exercises 1.9
Ek groundwater sample mein (mass ke hisaab se) 25 ppm nitrate ions, $N O_{3}^{-}$ ( $62 g m o l^{- 1}$ ), hain.
1. Ise mass percentage mein likho.
  %
  Solution dikhao
  $25 / 10^{6} \times 100 = 0.0025 %$ .
2. Nitrate ki molality nikaalo.
  mol/kg
  Solution dikhao
  $25 g$ yaani $25 / 62 = 0.403 m o l$ , practically $10^{6} g = 1000 k g$ paani mein, isliye $m = 4.0 \times 10^{- 4} m o l k g^{- 1}$ .
Short answerMediumNCERT Exercises 1.10
Hexane ko ethanol mein hilaya jaata hai. Mixing se pehle aur baad ke intermolecular attractions describe karo, aur Raoult's law se deviation ka type aur $Δ_{mix} H$ ka sign predict karo.
Solution dikhao
AnswerPositive deviation, $Δ_{mix} H > 0$ ke saath.
Pure ethanol ke molecules hydrogen bonds se jude hote hain; pure hexane mein sirf kamzor London (dispersion) forces hain. Milane par hexane molecules ethanol molecules ke beech aa jaate hain aur kai hydrogen bonds tod dete hain, badle mein sirf kamzor ethanol–hexane attractions dete hain.
Har kind ka molecule apne pure liquid se kam mazbooti se pakda jaata hai, isliye aasaani se nikalta hai: vapour pressure Raoult's-law value se upar hai (positive deviation). Hydrogen bonds todne ke liye energy chahiye, isliye mixing heat absorb karta hai ( $Δ_{mix} H > 0$ ) aur mixture thanda hota hai.
Short answerMediumNCERT Exercises 1.11
Garam, shaant dopahar mein ek chhichhle talaab ki machhliyan surface ke paas jama ho jaati hain. Gases ki solubility se iski wajah samjhao.
Solution dikhao
AnswerGaram paani mein kam oxygen ghuli hoti hai, kyunki gas ka ghulna exothermic hai.
Gas ka ghulna use condense karne jaisa hai, isliye heat release hoti hai ( $Δ_{sol} H < 0$ ). Le Chatelier's principle ke hisaab se temperature badhane par equilibrium $gas + water ⇌ solution$ wapas free gas ki taraf shift hota hai, isliye kam oxygen ghuli rehti hai ( $K_{H}$ temperature ke saath badhta hai).
Isliye garam talaab mein kam oxygen hoti hai, aur machhliyan surface layer mein aa jaati hain, jahan hawa se oxygen sabse tezi se bharti hai.
12
Short answerEasyNCERT Exercises 1.12
Henry's law batao aur use har observation samjhane ke liye use karo.
1. Henry's law batao.
  Solution dikhao
  AnswerConstant temperature par, solution ke upar gas ka partial pressure solution mein uske mole fraction ke proportional hota hai: $p = K_{H} x$ .
  Fixed temperature par $p = K_{H} x$ : liquid ke upar jitna zyada gas pressure, utni zyada gas ghulti hai.
2. Soda ki bottle kholne par fizz karti hai.
  Solution dikhao
  Use high CO2 pressure par seal kiya gaya tha, isliye bahut gas ghuli. Kholne par pressure hawa jitna gir jaata hai, isliye equilibrium mole fraction girta hai aur extra gas bubbles ban kar nikal jaati hai.
3. Divers ko dheere-dheere upar aana chahiye.
  Solution dikhao
  Gehraai par woh high pressure par hawa lete hain, isliye blood mein extra nitrogen ghulti hai. Jaldi upar aane se pressure achanak girta hai aur nitrogen blood mein bubbles banati hai (bends); dheere upar aane se woh lungs se nikal jaati hai.
4. Bahut oonchi chotiyon par climbers oxygen saath le jaate hain.
  Solution dikhao
  Oonchaai par oxygen ka partial pressure kam hota hai, isliye blood mein kam oxygen ghulti hai (anoxia); oxygen-enriched gas saans mein lene se uska partial pressure badhta hai.
NumericalEasyNCERT Exercises 1.13
Fixed temperature par, jab methane ka pressure $0.80 b a r$ hai to ek solvent ke sample mein $2.4 \times 10^{- 2} g$ methane ghulti hai. Utne hi solvent mein $6.0 \times 10^{- 2} g$ gholne ke liye kitna partial pressure chahiye?
bar
Solution dikhao
Henry's law se $p \propto x$ , aur chhoti amounts ke liye $x$ ghule mass ke proportional hai. Isliye $p = 0.80 \times (6.0 / 2.4) = 2.0 b a r$ .
14
Short answerMediumNCERT Exercises 1.14
Deviations aur heat of mixing.
1. Liquids X aur Y milane par heat absorb hoti hai aur volume thoda badhta hai. Raoult's law se tum kaun sa deviation expect karoge?
  Positive deviationNegative deviationKoi deviation nahi
  Answer dekho
  Answer (A) Positive deviation
  Heat absorb hone ka matlab X–Y attractions X–X aur Y–Y se kamzor hain: molecules aasaani se nikalte hain aur vapour pressure ideal se zyada hota hai.
2. Liquids U aur V milane par heat release hoti hai. Kaun sa statement fit hota hai?
  U–V attractions kamzor hain; vapour pressure Raoult se uparU–V attractions mazboot hain; vapour pressure Raoult se neecheMixture ideal hi hona chahiye
  Answer dekho
  Answer (B) U–V attractions mazboot hain; vapour pressure Raoult se neeche
  Heat release ( $Δ_{mix} H < 0$ ) matlab mazboot U–V attractions, kam escaping tendency aur negative deviation.
3. Ek-do sentences mein samjhao ki positive aur negative deviations ka kya matlab hai.
  Solution dikhao
  Positive deviation ka matlab naapa gaya vapour pressure Raoult's law ke prediction se zyada hai (A–B kamzor, $Δ_{mix} H > 0$ ); negative deviation ka matlab kam hai (A–B mazboot, $Δ_{mix} H < 0$ ).
NumericalHardNCERT Exercises 1.15
Ek non-volatile non-electrolyte ke 3.0 % (mass ke hisaab se) aqueous solution ka vapour pressure $373.15 K$ par $1.003 b a r$ hai, jahan pure paani ka $1.013 b a r$ hai. Solute ka molar mass nikaalo.
g/mol
Solution dikhao
Relative lowering: $x_{2} = (1.013 - 1.003) / 1.013 = 0.00987$ .
$100 g$ mein: $97.0 g$ paani $= 5.38 m o l$ . $x_{2} = n_{2} / (n_{1} + n_{2})$ se: $n_{2} = x_{2} n_{1} / (1 - x_{2}) = 0.0537 m o l$ .
$M_{2} = 3.0 / 0.0537 = 56 g m o l^{- 1}$ . (Dilute approximation $n_{2} \approx x_{2} n_{1}$ se $56.5$ aata hai, do significant figures tak same.)
16
NumericalMediumNCERT Exercises 1.16
Hexane ( $86 g m o l^{- 1}$ ) aur heptane ( $100 g m o l^{- 1}$ ) ideal solution banate hain. $333 K$ par unke vapour pressures $76.0 k P a$ aur $28.0 k P a$ hain. Ek mixture mein $43.0 g$ hexane aur $50.0 g$ heptane hai.
1. $333 K$ par iska vapour pressure kya hai?
  kPa
  Solution dikhao
  Har ek ke $n = 0.500 m o l$ , isliye $x = 0.500$ . $p = 0.500 \times 76.0 + 0.500 \times 28.0 = 52.0 k P a$ .
2. Vapour mein hexane ka mole fraction kya hai?
  Solution dikhao
  $y_{hexane} = 38.0 / 52.0 = 0.731$ .
NumericalMediumNCERT Exercises 1.17
Benzene ( $78 g m o l^{- 1}$ ) ka $298 K$ par vapour pressure $12.7 k P a$ hai. Isi temperature par benzene mein non-volatile solute ke $0.500 m o l k g^{- 1}$ solution ka vapour pressure calculate karo.
kPa
Solution dikhao
$1 k g$ benzene $= 1000 / 78 = 12.82 m o l$ , $0.500 m o l$ solute ke saath: $x_{2} = 0.500 / 13.32 = 0.0375$ .
$p = 12.7 \times (1 - 0.0375) = 12.2 k P a$ .
NumericalMediumNCERT Exercises 1.18
$92.0 g$ toluene ( $92 g m o l^{- 1}$ ) ka vapour pressure pure value ke 90.0 % tak kam karne ke liye usmein molar mass $60 g m o l^{- 1}$ wale non-volatile solute ka kitna mass gholna padega?
g
Solution dikhao
$p = x_{1} p^{\circ}$ , isliye $x_{1} = 0.900$ . Toluene: $n_{1} = 1.00 m o l$ .
$n_{2} = n_{1} (1 - x_{1}) / x_{1} = 1.00 \times 0.100 / 0.900 = 0.111 m o l$ , isliye mass $= 0.111 \times 60 = 6.67 g$ .
19
NumericalHardNCERT Exercises 1.19
$72.0 g$ paani mein $20.0 g$ non-volatile solute ka solution ek khaas temperature par $3.01 k P a$ vapour pressure dikhata hai. $36.0 g$ aur paani milane ke baad usi temperature par vapour pressure $3.07 k P a$ hai. (a) solute ka molar mass aur (b) is temperature par pure paani ka vapour pressure nikaalo.
1. Solute ka molar mass
  g/mol
  Solution dikhao
  Paani: $72.0 g = 4.00 m o l$ , phir $108 g = 6.00 m o l$ . Solute ko $n$ mol kaho.
  Pehla solution: $3.01 = 4.00 p^{\circ} / (4.00 + n)$ .
  Doosra solution: $3.07 = 6.00 p^{\circ} / (6.00 + n)$ .
  Pehle ko doosre se divide karne par $p^{\circ}$ hat jaata hai aur $n = 0.249 m o l$ milta hai, isliye $M = 20.0 / 0.249 = 80 g m o l^{- 1}$ .
2. Pure paani ka vapour pressure
  kPa
  Solution dikhao
  $p^{\circ} = 3.01 \times (4.00 + 0.249) / 4.00 = 3.20 k P a$ .
NumericalMediumNCERT Exercises 1.20
4.0 % (mass ke hisaab se) aqueous urea solution ( $60 g m o l^{- 1}$ ) $271.86 K$ par jamta hai. $K_{f}$ use kiye bina, 4.0 % aqueous glucose solution ( $180 g m o l^{- 1}$ ) ka freezing point nikaalo. Pure paani $273.15 K$ par jamta hai.
K
Solution dikhao
Dono solutions mein $96.0 g$ paani mein $4.0 g$ solute hai, isliye unki molalities molar masses ke inverse ratio mein hain, $60 : 180 = 1 : 3$ .
$Δ T_{f} (urea) = 273.15 - 271.86 = 1.29 K$ , isliye $Δ T_{f} (glucose) = 1.29 / 3 = 0.43 K$ aur glucose solution $272.72 K$ par jamta hai.
NumericalHardNCERT Exercises 1.21
Elements X aur Y ke do compounds ke formulae $X Y_{3}$ aur $X Y_{5}$ hain. $25.0 g$ benzene ( $K_{f} = 5.12 K k g m o l^{- 1}$ ) mein ghol kar, $1.00 g$ $X Y_{3}$ freezing point $1.49 K$ kam karta hai aur $1.00 g$ $X Y_{5}$ $0.984 K$ kam karta hai. X aur Y ke atomic masses nikaalo, aur batao ki elements kya ho sakte hain.
Solution dikhao
Answer $M (X Y_{3}) = 137$ , $M (X Y_{5}) = 208$ ; Y ≈ 35.3 (chlorine), X ≈ 31.4 (phosphorus).
$M (X Y_{3}) = 5.12 \times 1.00 \times 1000 / (1.49 \times 25.0) = 137.4 g m o l^{- 1}$ ; $M (X Y_{5}) = 5.12 \times 1000 / (0.984 \times 25.0) = 208.1 g m o l^{- 1}$ .
$2 Y = 208.1 - 137.4 = 70.7$ , isliye $Y = 35.3$ . Phir $X = 137.4 - 3 \times 35.3 = 31.4$ .
Yeh chlorine (35.5) aur phosphorus (31.0) se fit hote hain: compounds $P C l_{3}$ aur $P C l_{5}$ hain.
NumericalMediumNCERT Exercises 1.22
$310 K$ par per litre $18.0 g$ glucose wale solution ka osmotic pressure $2.58 b a r$ hai. Usi temperature par kaun si glucose concentration $7.74 b a r$ degi, jo blood ke osmotic pressure ke kareeb hai?
M
Solution dikhao
$18.0 g L^{- 1}$ glucose $0.100 M$ hai. Kyunki $Π = C R T$ , $C_{2} = C_{1} Π_{2} / Π_{1} = 0.100 \times 7.74 / 2.58 = 0.300 M$ ( $54.0 g L^{- 1}$ ).
23
Quick answerMediumNCERT Exercises 1.23
Substances ke har pair ko unke beech ke sabse important attraction se match karo.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
pentane aur hexane : London (dispersion) forces
potassium bromide aur paani : ion–dipole attraction
ethanol aur paani : hydrogen bonding
acetone aur ethyl ethanoate : dipole–dipole attraction
Alkanes non-polar hain; ions polar water molecules ki taraf attract hote hain; ethanol aur paani dono mein O–H groups hain; acetone aur ethyl ethanoate polar hain lekin unmein O–H ya N–H nahi hai, isliye woh aapas mein hydrogen bond nahi bana sakte.
24
Quick answerMediumNCERT Exercises 1.24
Inhe hexane mein badhti solubility ke order mein lagao: pentane, sodium bromide, acetone, ethylene glycol.
1. acetone
2. ethylene glycol
3. pentane
4. sodium bromide
Answer dekho
Answer
sodium bromide
ethylene glycol
acetone
pentane
Ionic solid non-polar liquid mein mushkil se ghulta hai. Ethylene glycol, do O–H groups ke saath, strongly hydrogen-bonded hai aur bas thoda soluble hai. Acetone polar hai lekin hexane ke saath mix ho jaata hai. Pentane, hexane jaisa non-polar alkane, har proportion mein mix hota hai.
25
Quick answerMediumNCERT Exercises 1.25
Har compound ko room temperature par paani mein uski solubility se match karo.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
hexane : insoluble
1-butanol : partially soluble
glycerol : highly soluble (miscible)
Hexane mein koi polar group nahi hai. 1-Butanol ka O–H uska kuch hissa ghol deta hai (per 100 g paani lagbhag 8 g), lekin chaar-carbon chain ise limit karti hai. Glycerol ke chhote molecule par teen O–H groups hain aur woh poori tarah mix hota hai.
NumericalHardNCERT Exercises 1.26
Brine ke ek sample mein har $1.00 k g$ paani ke liye $69.0 g$ $N a^{+}$ ions hain (matching chloride ke saath, NaCl ki tarah). Uski density $1.11 g m L^{- 1}$ hai. $N a^{+}$ ki molarity calculate karo.
M
Solution dikhao
$n (N a^{+}) = 69.0 / 22.99 = 3.00 m o l$ , saath mein $3.00 \times 35.45 = 106.4 g$ $C l^{-}$ .
Brine ka mass $= 1000 + 69.0 + 106.4 = 1175 g$ ; volume $= 1175 / 1.11 = 1059 m L$ .
$N a^{+}$ ki molarity $= 3.00 / 1.059 = 2.83 M$ .
NumericalMediumNCERT Exercises 1.27
$298 K$ par silver bromide, $A g B r$ , ka solubility product $5.0 \times 10^{- 13}$ hai. Is temperature par paani mein AgBr ki sabse badi molarity kitni ho sakti hai?
M
Solution dikhao
Ghulne wali har AgBr unit ek $A g^{+}$ aur ek $B r^{-}$ ion deti hai. Solubility $s$ ke saath $[A g^{+}] = [B r^{-}] = s$ aur $K_{s p} = s^{2}$ .
$s = \sqrt{5.0 \times 10^{- 13}} = 7.1 \times 10^{- 7} m o l L^{- 1}$ (per litre lagbhag $0.13 m g$ ).
NumericalEasyNCERT Exercises 1.28
$75 g$ ethanol mein $2.5 g$ paracetamol ( $C_{8} H_{9} N O_{2}$ ) ghola jaata hai. Paracetamol ka mass percentage calculate karo.
%
Solution dikhao
$2.5 / (2.5 + 75) \times 100 = 3.23 %$ .
NumericalMediumNCERT Exercises 1.29
Ek patient ko $2.0 \times 10^{- 3} m o l k g^{- 1}$ aqueous solution se $50 m g$ caffeine ( $C_{8} H_{10} N_{4} O_{2}$ , $194.2 g m o l^{- 1}$ ) deni hai. Solution ka kitna mass chahiye?
g
Solution dikhao
$n = 0.050 / 194.2 = 2.57 \times 10^{- 4} m o l$ . Zaroori paani $= n / m = 2.57 \times 10^{- 4} / 2.0 \times 10^{- 3} = 0.129 k g$ .
$0.05 g$ caffeine jodne se is precision par kuch nahi badalta: lagbhag $129 g$ solution.
NumericalEasyNCERT Exercises 1.30
Ethanol mein $0.120 M$ solution ka $500 m L$ banane ke liye kitna salicylic acid ( $C_{7} H_{6} O_{3}$ , $138.1 g m o l^{- 1}$ ) chahiye?
g
Solution dikhao
$n = 0.120 \times 0.500 = 0.0600 m o l$ ; mass $= 0.0600 \times 138.1 = 8.29 g$ .
31
Quick answerMediumNCERT Exercises 1.31
Ethanoic acid, chloroethanoic acid aur dichloroethanoic acid ki barabar molalities paani mein gholi jaati hain. Inhe sabse chhote se sabse bade freezing-point depression ke order mein lagao.
1. chloroethanoic acid
2. ethanoic acid
3. dichloroethanoic acid
Answer dekho
Answer
ethanoic acid
chloroethanoic acid
dichloroethanoic acid
Har chlorine –COOH group se electron density door kheenchta hai aur anion ko stabilise karta hai, isliye chlorines ki sankhya ke saath acid strength badhti hai. Strong acid zyada dissociate hota hai, jisse bada $i$ aur bada $Δ T_{f}$ milta hai.
NumericalHardNCERT Exercises 1.32
$400 g$ paani mein $8.00 g$ chloroethanoic acid ( $C l C H_{2} C O O H$ , $94.5 g m o l^{- 1}$ , $K_{a} = 1.35 \times 10^{- 3}$ ) ghola jaata hai. Freezing point ke depression ka estimate lagao ( $K_{f} = 1.86 K k g m o l^{- 1}$ ).
K
Solution dikhao
$m = (8.00 / 94.5) / 0.400 = 0.212 m o l k g^{- 1}$ ; $c \approx 0.212 M$ lo.
$0.212 α^{2} + 1.35 \times 10^{- 3} α - 1.35 \times 10^{- 3} = 0$ solve karo: $α = 0.077$ , isliye $i = 1 + α = 1.077$ .
$Δ T_{f} = i K_{f} m = 1.077 \times 1.86 \times 0.212 = 0.424 K$ (acid dissociate na hota to $0.394 K$ ).
33
NumericalHardNCERT Exercises 1.33
$500 g$ paani mein $13.9 g$ bromoethanoic acid ( $B r C H_{2} C O O H$ , $138.9 g m o l^{- 1}$ ) ghola jaata hai, aur freezing point $0.401 K$ girta hai. Acid ke liye $i$ nikaalo, aur usse acid ka $K_{a}$ ( $K_{f} = 1.86 K k g m o l^{- 1}$ ).
1. van't Hoff factor
  Solution dikhao
  $13.9 g$ yaani $0.100 m o l$ , isliye $m = 0.200 m o l k g^{- 1}$ .
  Calculated $Δ T = 1.86 \times 0.200 = 0.372 K$ , isliye $i = 0.401 / 0.372 = 1.078$ .
2. Dissociation constant
  Solution dikhao
  $α = i - 1 = 0.078$ ; $K_{a} = 0.200 \times {0.078}^{2} / (1 - 0.078) = 1.3 \times 10^{- 3}$ .
NumericalMediumNCERT Exercises 1.34
$308 K$ par paani ka vapour pressure $42.2 m m H g$ hai. Is temperature par $500 g$ paani mein $36.0 g$ glucose ( $180 g m o l^{- 1}$ ) ke solution ka vapour pressure calculate karo.
mmHg
Solution dikhao
$n_{2} = 0.200 m o l$ , $n_{1} = 500 / 18.02 = 27.75 m o l$ ; $x_{1} = 27.75 / 27.95 = 0.9928$ .
$p = 0.9928 \times 42.2 = 41.9 m m H g$ .
35
NumericalMediumNCERT Exercises 1.35
$298 K$ par paani mein nitrogen ka Henry's law constant $6.5 \times 10^{7} m m H g$ hai. $593 m m H g$ partial pressure (sea level par hawa mein uska hissa) par nitrogen ke sampark mein paani mein nitrogen ka mole fraction aur molality nikaalo.
1. Mole fraction
  Solution dikhao
  $x = 593 / 6.5 \times 10^{7} = 9.1 \times 10^{- 6}$ .
2. Molality
  mol/kg
  Solution dikhao
  Per kilogram paani $n \approx x \times 55.5 = 5.1 \times 10^{- 4} m o l$ .
36
NumericalMediumNCERT Exercises 1.36
$50.0 g$ liquid P ( $100 g m o l^{- 1}$ ) ko $400 g$ liquid Q ( $200 g m o l^{- 1}$ ) ke saath mila kar ideal solution banaya jaata hai. Pure Q ka vapour pressure $300 t o r r$ hai, aur solution ka total vapour pressure $280 t o r r$ hai. Pure P ka vapour pressure aur solution mein uska partial pressure nikaalo.
1. Solution mein P ka partial pressure
  torr
  Solution dikhao
  $n_{P} = 0.500$ , $n_{Q} = 2.00$ , isliye $x_{P} = 0.200$ , $x_{Q} = 0.800$ . $p_{Q} = 0.800 \times 300 = 240 t o r r$ , isliye $p_{P} = 280 - 240 = 40 t o r r$ .
2. Pure P ka vapour pressure
  torr
  Solution dikhao
  $p_{P}^{\circ} = p_{P} / x_{P} = 40 / 0.200 = 200 t o r r$ .
DiagramHardNCERT Exercises 1.37
Pure liquids P aur Q ke $320 K$ par vapour pressures $520$ aur $380 m m H g$ hain. Unke mixtures ke upar naape gaye partial pressures neeche diye gaye hain. $p_{P}$ , $p_{Q}$ aur $p_{total}$ ko $x_{P}$ ke against plot karo, ideal (Raoult's law) lines jodo, aur tay karo ki mixture kis kind ka deviation dikhata hai.
Solution dikhao
AnswerPositive deviation: har naapa gaya total ideal line ke upar hai; $x_{P} = 0.8$ ke paas total 520 mmHg se bhi upar chala jaata hai, isliye wahan ek minimum-boiling azeotrope banta hai.
Table 1.8320 K par naape gaye pressures (mmHg), comparison ke liye ideal total ke saath
$x_{P}$ $p_{P}$ $p_{Q}$ total ideal
0 0 380 380 380
0.2 153 311 464 408
0.4 258 251 509 436
0.6 343 189 532 464
0.8 426 112 538 492
1.0 520 0 520 520
Ideal values $p_{P} = 520 x_{P}$ aur $p_{Q} = 380 (1 - x_{P})$ se aati hain. Har naapa gaya partial pressure aur total apni ideal value se upar hai, isliye mixture positive deviation dikhata hai: P–Q attractions P–P aur Q–Q se kamzor hain.
$x_{P} = 0.8$ ke paas naapa gaya total ( $538 m m H g$ ) pure P ke $520 m m H g$ se bhi zyada hai, isliye total curve mein ek maximum hai. Lagbhag is composition ka mixture minimum-boiling azeotrope hai.
NumericalMediumNCERT Exercises 1.38
Toluene ( $92 g m o l^{- 1}$ ) aur ethylbenzene ( $106 g m o l^{- 1}$ ) ideal solution banate hain. $360 K$ par unke vapour pressures $48.9 k P a$ aur $21.6 k P a$ hain. $46.0 g$ toluene aur $106 g$ ethylbenzene ke mixture ke upar vapour mein toluene ka mole fraction nikaalo.
Solution dikhao
$n_{tol} = 0.500 m o l$ , $n_{eb} = 1.00 m o l$ , isliye $x_{tol} = 0.333$ .
$p_{tol} = 0.333 \times 48.9 = 16.3 k P a$ ; $p_{eb} = 0.667 \times 21.6 = 14.4 k P a$ ; total $30.7 k P a$ .
$y_{tol} = 16.3 / 30.7 = 0.531$ : liquid ka ek-tihaai lekin vapour ka aadhe se zyada.
39
NumericalHardNCERT Exercises 1.39
Hawa volume ke hisaab se lagbhag 21 % oxygen aur 78 % nitrogen hai. $298 K$ par paani $5.00 b a r$ ke total pressure par hawa ke saath equilibrium mein hai. $O_{2}$ ke liye $K_{H} = 44 k b a r$ aur $N_{2}$ ke liye $86 k b a r$ use karke, har ghuli gas ka mole fraction nikaalo. Kya ghuli hui hawa upar ki hawa se oxygen mein richer hai ya poorer?
1. Oxygen ka mole fraction
  Solution dikhao
  $p (O_{2}) = 0.21 \times 5.00 = 1.05 b a r$ ; $x = 1.05 / 44 000 = 2.39 \times 10^{- 5}$ .
2. Nitrogen ka mole fraction
  Solution dikhao
  $p (N_{2}) = 0.78 \times 5.00 = 3.90 b a r$ ; $x = 3.90 / 86 000 = 4.53 \times 10^{- 5}$ .
3. Oxygen mein richer ya poorer?
  Solution dikhao
  AnswerRicher.
  Ghuli N2 : O2 $= 4.53 / 2.39 = 1.9$ , jabki hawa mein $78 / 21 = 3.7$ . Oxygen zyada soluble gas hai (chhota $K_{H}$ ), isliye ghuli hui hawa atmosphere se lagbhag do guna oxygen-rich hoti hai, jo aquatic animals ki madad karta hai.
NumericalMediumNCERT Exercises 1.40
$300 K$ par $1.20 b a r$ osmotic pressure wala $1.50 L$ solution banane ke liye paani mein kitna magnesium chloride ( $M g C l_{2}$ , $95.2 g m o l^{- 1}$ , $i = 2.60$ ) gholna padega? ( $R = 0.0831 L b a r m o l^{- 1} K^{- 1}$ )
g
Solution dikhao
$n = Π V / (i R T) = 1.20 \times 1.50 / (2.60 \times 0.0831 \times 300) = 0.0278 m o l$ .
Mass $= 0.0278 \times 95.2 = 2.64 g$ .
NumericalMediumNCERT Exercises 1.41
$500 m L$ paani mein $40.0 m g$ sodium sulfate ( $N a_{2} S O_{4}$ , $142.0 g m o l^{- 1}$ ) ke solution ka $37^{\circ} C$ par osmotic pressure nikaalo, poora dissociation maan kar. ( $R = 0.0831 L b a r m o l^{- 1} K^{- 1}$ )
bar
Solution dikhao
Har $N a_{2} S O_{4}$ do $N a^{+}$ ions aur ek $S O_{4}^{2 -}$ ion deta hai, isliye $i = 3$ .
$0.500 L$ mein $n = 0.0400 / 142.0 = 2.82 \times 10^{- 4} m o l$ : $C = 5.63 \times 10^{- 4} M$ .
$Π = i C R T = 3 \times 5.63 \times 10^{- 4} \times 0.0831 \times 310 = 0.0436 b a r$ ( $4.36 k P a$ ).
Quick answerEasy
Inmein se kaun si colligative property nahi hai?
Osmotic pressureElevation of boiling pointPure solvent ka vapour pressureDepression of freezing point
Answer dekho
Answer (C) Pure solvent ka vapour pressure
Pure solvent ka vapour pressure sirf solvent ki property hai; colligative property solute ki wajah se vapour pressure ka lowering hai.
Quick answerMedium
Ek salt ka $0.100 m o l k g^{- 1}$ aqueous solution $- 0.502^{\circ} C$ par jamta hai. Iska van't Hoff factor kya hai? ( $K_{f} = 1.86 K k g m o l^{- 1}$ )
Answer dekho
Answer 2.7
Calculated $Δ T_{f} = 0.186 K$ ; $i = 0.502 / 0.186 = 2.7$ , jo $C a C l_{2}$ jaisa salt suggest karta hai jo poori tarah dissociate nahi hua.
Quick answerEasy
Same temperature par same osmotic pressure wale do solutions ko solutions kehte hain.
Answer dekho
Answer isotonic
Isotonic solutions semipermeable membrane ke paar koi net osmosis nahi dikhate.
45
Quick answerMedium
In aqueous solutions ko, sab $0.01 m o l k g^{- 1}$ , badhte boiling point ke order mein lagao.
1. $K N O_{3}$
2. $N a_{2} S O_{4}$
3. glucose
4. $A l C l_{3}$
Answer dekho
Answer
glucose
$K N O_{3}$
$N a_{2} S O_{4}$
$A l C l_{3}$
Ideal $i$ values 1, 2, 3 aur 4 hain, isliye boiling-point elevation, aur boiling point, isi order mein badhte hain.
Quick answerEasy
Solution garam karne par uski molality badal jaati hai.
SahiGalat
Answer dekho
Answer Galat
False. Molality per kilogram solvent solute ke moles hai; dono mein se koi temperature ke saath nahi badalta. Molarity, jo volume use karti hai, badalti hai.

$x_{P}$	$p_{P}$	$p_{Q}$	total	ideal
0	0	380	380	380
0.2	153	311	464	408
0.4	258	251	509	436
0.6	343	189	532	464
0.8	426	112	538	492
1.0	520	0	520	520

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Intext Question 1.1	Practice: concentration ke sawaal (Intext Questions 1.1–1.5) · Q1
NCERT Intext Question 1.2	Practice: concentration ke sawaal (Intext Questions 1.1–1.5) · Q2
NCERT Intext Question 1.3	Practice: concentration ke sawaal (Intext Questions 1.1–1.5) · Q3
NCERT Intext Question 1.4	Practice: concentration ke sawaal (Intext Questions 1.1–1.5) · Q4
NCERT Intext Question 1.5	Practice: concentration ke sawaal (Intext Questions 1.1–1.5) · Q5
NCERT Intext Question 1.6	Practice: Henry's law ke sawaal (Intext Questions 1.6–1.7) · Q1
NCERT Intext Question 1.7	Practice: Henry's law ke sawaal (Intext Questions 1.6–1.7) · Q2
NCERT Intext Question 1.8	Practice: liquid mixtures ke sawaal (Intext Question 1.8) · Q1
NCERT Intext Question 1.9	Practice: colligative properties ke sawaal (Intext Questions 1.9–1.12) · Q1
NCERT Intext Question 1.10	Practice: colligative properties ke sawaal (Intext Questions 1.9–1.12) · Q2
NCERT Intext Question 1.11	Practice: colligative properties ke sawaal (Intext Questions 1.9–1.12) · Q3
NCERT Intext Question 1.12	Practice: colligative properties ke sawaal (Intext Questions 1.9–1.12) · Q4
NCERT Exercises 1.1	Exercises · Q1
NCERT Exercises 1.2	Exercises · Q2
NCERT Exercises 1.3	Exercises · Q3
NCERT Exercises 1.4	Exercises · Q4
NCERT Exercises 1.5	Exercises · Q5
NCERT Exercises 1.6	Exercises · Q6
NCERT Exercises 1.7	Exercises · Q7
NCERT Exercises 1.8	Exercises · Q8
NCERT Exercises 1.9	Exercises · Q9
NCERT Exercises 1.10	Exercises · Q10
NCERT Exercises 1.11	Exercises · Q11
NCERT Exercises 1.12	Exercises · Q12
NCERT Exercises 1.13	Exercises · Q13
NCERT Exercises 1.14	Exercises · Q14
NCERT Exercises 1.15	Exercises · Q15
NCERT Exercises 1.16	Exercises · Q16
NCERT Exercises 1.17	Exercises · Q17
NCERT Exercises 1.18	Exercises · Q18
NCERT Exercises 1.19	Exercises · Q19
NCERT Exercises 1.20	Exercises · Q20
NCERT Exercises 1.21	Exercises · Q21
NCERT Exercises 1.22	Exercises · Q22
NCERT Exercises 1.23	Exercises · Q23
NCERT Exercises 1.24	Exercises · Q24
NCERT Exercises 1.25	Exercises · Q25
NCERT Exercises 1.26	Exercises · Q26
NCERT Exercises 1.27	Exercises · Q27
NCERT Exercises 1.28	Exercises · Q28
NCERT Exercises 1.29	Exercises · Q29
NCERT Exercises 1.30	Exercises · Q30
NCERT Exercises 1.31	Exercises · Q31
NCERT Exercises 1.32	Exercises · Q32
NCERT Exercises 1.33	Exercises · Q33
NCERT Exercises 1.34	Exercises · Q34
NCERT Exercises 1.35	Exercises · Q35
NCERT Exercises 1.36	Exercises · Q36
NCERT Exercises 1.37	Exercises · Q37
NCERT Exercises 1.38	Exercises · Q38
NCERT Exercises 1.39	Exercises · Q39
NCERT Exercises 1.40	Exercises · Q40
NCERT Exercises 1.41	Exercises · Q41

Quick check: MCQs

1.1 Types of Solutions3 questions
Mole fraction, molarity aur molality4 questions
1.3.2 Solubility of a Gas in a Liquid4 questions
1.4.3 Vapour Pressure of Solutions of Solids in Liquids4 questions
1.5.2 Non-ideal Solutions3 questions
1.6.3 Depression of Freezing Point4 questions
1.6.5 Reverse Osmosis and Water Purification4 questions
1.7 Abnormal Molar Masses4 questions

Glossary · 15

Key terms (English + Hindi)

azeotropeस्थिरक्वाथीsthirkwathi: Aisa liquid mixture jo constant temperature par ubal kar same composition ka vapour deta hai, isliye distillation use alag nahi kar sakta.Chapter mein pehli baar yahan aaya →

colligative propertyअणुसंख्य गुणधर्मanusankhya gundharm: Dilute solution ki aisi property jo solute particles ki sankhya par depend karti hai, is par nahi ki woh kya hain.Chapter mein pehli baar yahan aaya →

Henry's lawहेनरी का नियमHenry ka niyam: Constant temperature par solution ke upar gas ka partial pressure solution mein uske mole fraction ke proportional hota hai, $p = K_{H} x$ .

ideal solutionआदर्श विलयनaadarsh vilayan: Aisa solution jo har composition par Raoult's law follow kare, aur mixing par na heat change ho na volume change.Chapter mein pehli baar yahan aaya →

molalityमोललताmolalta: Per kilogram solvent solute ke moles, $m o l k g^{- 1}$ (m) mein; yeh temperature par depend nahi karti.Chapter mein pehli baar yahan aaya →
molarityमोलरताmolarta: Per litre solution solute ke moles, $m o l L^{- 1}$ (M) mein; yeh temperature ke saath thoda badalti hai.Chapter mein pehli baar yahan aaya →
mole fractionमोल अंशmol ansh: Ek component ke moles ko saare components ke total moles se divide karke; solution ke mole fractions ka total 1 hota hai.Chapter mein pehli baar yahan aaya →

osmotic pressureपरासरण दाबparasaran daab: Woh extra pressure jo solution par lagana padta hai taaki semipermeable membrane se hokar usmein solvent na aaye; $Π = C R T$ .Chapter mein pehli baar yahan aaya →

Raoult's lawराउल्ट का नियमRaoult ka niyam: Solution ke har volatile component ka partial vapour pressure uske mole fraction guna uske pure vapour pressure ke barabar hota hai, $p_{i} = x_{i} p_{i}^{\circ}$ .Chapter mein pehli baar yahan aaya →

saturated solutionसंतृप्त विलयनsantript vilayan: Bina ghule solute ke saath dynamic equilibrium mein solution, jo us temperature aur pressure par aur nahi ghol sakta.Chapter mein pehli baar yahan aaya →
solubilityविलेयताvileyata: Kisi substance ki woh sabse badi amount jo diye gaye temperature aur pressure par solvent ki batayi gayi amount mein ghulti hai.Chapter mein pehli baar yahan aaya →
soluteविलेयvilay: Solution ka solvent ke alawa koi bhi component.Chapter mein pehli baar yahan aaya →
solutionविलयनvilayan: Do ya zyada components se bana aisa mixture jo molecules ke scale tak har jagah uniform ho.Chapter mein pehli baar yahan aaya →
solventविलायकvilaayak: Solution ka woh component jo aam taur par sabse zyada amount mein hota hai aur uski physical state tay karta hai.Chapter mein pehli baar yahan aaya →

van't Hoff factorवान्ट हॉफ गुणकvant Hoff gunak: Sach mein maujood particles aur ghuli formula units ka ratio $i$ ; dissociation mein 1 se zyada, association mein 1 se kam.Chapter mein pehli baar yahan aaya →

$x_{P}$	$p_{P}$	$p_{Q}$	total	ideal
0	0	380	380	380
0.2	153	311	464	408
0.4	258	251	509	436
0.6	343	189	532	464
0.8	426	112	538	492
1.0	520	0	520	520

$x_{P}$	$p_{P}$	$p_{Q}$	total	ideal
0	0	380	380	380
0.2	153	311	464	408
0.4	258	251	509	436
0.6	343	189	532	464
0.8	426	112	538	492
1.0	520	0	520	520

1.1 Types of Solutions

1.2 Expressing Concentration of Solutions

Percentages aur parts per million

Mole fraction, molarity aur molality

1.3 Solubility

1.3.1 Solubility of a Solid in a Liquid

1.3.2 Solubility of a Gas in a Liquid

1.4 Vapour Pressure of Liquid Solutions

1.4.1 Vapour Pressure of Liquid-Liquid Solutions

1.4.2 Raoult's Law as a special case of Henry's Law

1.4.3 Vapour Pressure of Solutions of Solids in Liquids

1.5 Ideal and Non-ideal Solutions

1.5.1 Ideal Solutions

1.5.2 Non-ideal Solutions

1.6 Colligative Properties and Determination of Molar Mass

1.6.1 Relative Lowering of Vapour Pressure

1.6.2 Elevation of Boiling Point

1.6.3 Depression of Freezing Point

1.6.4 Osmosis and Osmotic Pressure

1.6.5 Reverse Osmosis and Water Purification

1.7 Abnormal Molar Masses

Apna NCERT exercise question dhoondo

Quick check: MCQs

Tum chapter ke end tak pahunch gaye

$x_{P}$	$p_{P}$	$p_{Q}$	total	ideal
0	0	380	380	380
0.2	153	311	464	408
0.4	258	251	509	436
0.6	343	189	532	464
0.8	426	112	538	492
1.0	520	0	520	520