Solutions: Class 12 Chemistry · Chapter 1

Chemistry Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 11 Oct 2026 · Updated 11 Oct 2026

115 min
1 simulation
58 practice questions
30 MCQs

What you will learn in Solutions

Classify binary solutions by the physical states of solute and solvent and give an example of each type
Express the concentration of a solution as mass, volume or mass-by-volume percentage, ppm, mole fraction, molarity and molality, convert between them, and say which change with temperature
Explain how the nature of solute and solvent, temperature and pressure affect solubility, and use Henry's law for gases
Apply Raoult's law to mixtures of two volatile liquids and to solutions of non-volatile solutes, including the composition of the vapour
Distinguish ideal from non-ideal solutions, link positive and negative deviations to intermolecular forces, and describe azeotropes
Calculate the four colligative properties and use them to find the molar mass of a solute
Explain osmosis, osmotic pressure, isotonic, hypertonic and hypotonic solutions, and reverse osmosis
Use the van't Hoff factor to correct for dissociation and association and to find abnormal molar masses and degrees of dissociation or association

Original companion text aligned to NCERT Chemistry Class 12, Chapter 1, pp. 1–30. Not affiliated with or endorsed by NCERT.

A packet of oral rehydration salts tells you to empty it into exactly one litre of clean water. Use half a litre and the drink becomes too concentrated: instead of rehydrating a sick child it can pull water out of the gut. Jewellers stamp "22K" on gold that is about 91.6 % gold by mass, the rest being copper and silver. Petrol pumps across India now sell E20, petrol with 20 % ethanol by volume. In each case a mixture's behaviour depends on exactly how much of each ingredient it holds.

This unit is about such mixtures, mainly liquid ones. We first sort solutions into types and learn the different ways chemists state their concentration. Then we ask how much of a solid or a gas will dissolve, what happens to the vapour above a liquid mixture, and why some mixtures behave 'ideally' while others do not. Finally we meet the colligative properties, which depend only on how many solute particles are present, and use them to weigh molecules too small to see.

1.1 Types of Solutions

Stir a spoonful of sugar into water and, after a minute, every sip tastes equally sweet. A solution (विलयन) is exactly this: a homogeneous mixture, one whose composition and properties are the same everywhere in it, down to the scale of molecules. Muddy water fails the test: the silt settles and the bottom of the glass differs from the top.

We name the parts of a solution by their role. The solvent is normally the component present in the largest amount, and it fixes the physical state of the whole solution: salt water is a liquid because water is. Every other component is a solute. In this unit we deal almost entirely with binary solutions, which have just two components: one solvent and one solute.

Definition

Solution

A homogeneous mixture of two or more components; the component that sets its physical state (usually the one in largest amount) is the solvent, and the others are solutes.

ExampleAir in a sealed room, sugar syrup, and the alloy in a steel spoon are all solutions.

Each of the two components may be gaseous, liquid or solid, which makes nine possible kinds of binary solution. Whatever the solute, the solvent's state decides whether we call the solution gaseous, liquid or solid (Fig. 1.1), and Table 1.1 gives one example of each kind.

4/4Three states, one idea: the mixture is uniform at the scale of particles.

Fig. 1.1In every solution the solute particles are spread evenly among the solvent particles; the solvent decides whether that happens in a gas, a liquid or a solid.NCERT p. 2

Three panels show solutions at the scale of particles, with circles for solvent particles and squares for solute particles. In step 1 the gaseous solution is highlighted: a few particles far apart in a closed jar, labelled humid air. In step 2 the liquid solution is highlighted: particles crowded together but jumbled, in a beaker, labelled salt water. In step 3 the solid solution is highlighted: particles in a regular grid, with a few grid places taken by solute particles, labelled steel. In step 4 all three panels are shown together; in each, the solute particles are spread evenly among the solvent particles.

Table 1.1Nine kinds of binary solution, grouped by the state of the solventNCERT Table 1.1 · p. 2

Solution	Solute	Solvent	Example
Gaseous	gas	gas	argon mixed into the neon of a glowing sign tube
Gaseous	liquid	gas	water vapour in humid air on a monsoon day
Gaseous	solid	gas	naphthalene (mothball) vapour in the air of a closed cupboard
Liquid	gas	liquid	carbon dioxide in soda water
Liquid	liquid	liquid	ethanoic (acetic) acid in water, as vinegar
Liquid	solid	liquid	sugar and salt in nimbu-pani
Solid	gas	solid	hydrogen stored inside a metal-alloy canister (a metal hydride)
Solid	liquid	solid	mercury in a silver–tin alloy, as dental amalgam
Solid	solid	solid	chromium and nickel in iron, as stainless steel

1.2 Expressing Concentration of Solutions

We can call a cup of tea 'weak' or 'strong', and a syrup 'thin' or 'thick', but such words mean different things to different people. A pharmacist preparing a drip, or an engineer blending fuel, needs a number. The concentration of a solution says how much solute is present in a given amount of solvent or solution, and chemists use several measures of it. Each suits a different job, so learn them all and practise converting between them.

Fig. 1.2 follows one solution, $6.0 g$ of urea dissolved in $100 g$ of water, through the measures described below.

6/6At $60^{\circ} C$ the liquid expands: molarity drops, molality does not.

Fig. 1.2One solution, five ways to state its concentration. Only the measure built on volume changes when the solution is warmed.NCERT p. 2

A beaker on the left holds a solution; a card on the right shows one measure of its concentration per step. Step 1 shows the recipe: 6.0 g of urea plus 100 g of water gives 106 g of solution. Step 2 shows the mass percentage, 6.0 divided by 106 times 100, about 5.7 percent. Step 3 shows the amounts in moles, 0.10 mol of urea and 5.55 mol of water, giving a mole fraction of urea of 0.018. Step 4 shows the molality: 0.10 mol per 0.100 kg of water, or 1.0 mol per kg. Step 5 shows the molarity: the solution's volume is 104 mL, so 0.96 mol per litre. Step 6 shows the beaker warmed to 60 degrees Celsius: the liquid level rises slightly, the molarity falls to 0.94 mol per litre, while the molality stays 1.0 mol per kg.

Percentages and parts per million

(i) Mass percentage (w/w). This compares masses only, so it is easy to measure with a balance and is common in industry and on product labels. A brine for lime pickle made from $12 g$ of salt and $88 g$ of water is 12 % salt by mass.

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

(1.1)

where

$m_{A}$: mass of component A in the solution
$m_{solution}$: total mass of the solution

(ii) Volume percentage (V/V). For solutions made by mixing liquids it is often easier to measure volumes. A hand sanitiser labelled '70 % isopropyl alcohol (v/v)' contains $70 m L$ of the alcohol in every $100 m L$ of the finished liquid. Note that the divisor is the volume of the solution: liquids do not always add up exactly in volume when mixed.

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

(1.2)

where

$V_{A}$: volume of component A
$V_{solution}$: total volume of the solution

(iii) Mass by volume percentage (w/V). Hospitals and pharmacies prefer grams of solute per $100 m L$ of solution. A '5 % dextrose' drip holds $5 g$ of glucose in each $100 m L$ of the fluid.

(iv) Parts per million (ppm). For traces, percentages give awkward strings of zeros, so we count parts per million parts. Like percentages, ppm can be mass/mass, volume/volume or mass/volume. Carbon dioxide makes up about 420 ppm of today's air by volume, and $1 m g$ of a pesticide in $1 k g$ of water is 1 ppm by mass. Water-quality and air-pollution limits are usually written this way, or in $m g L^{- 1}$ , which for dilute water samples is almost the same as ppm.

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

(1.3)

Mole fraction, molarity and molality

(v) Mole fraction. Chemistry happens molecule by molecule, so many laws of solutions are written in moles. The mole fraction of a component, written $x$ with the component as a subscript, is its share of all the moles present.

x_{A} = \frac{mol of A}{total mol}

(1.4)

For a binary solution of A and B containing $n_{A}$ and $n_{B}$ moles, and for a solution with any number of components:

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)

Because every mole is counted once in the denominator, the mole fractions of all the components add up to exactly 1. In a binary solution, once you know $x_{A}$ you also know $x_{B} = 1 - x_{A}$ . Mole fractions are the natural language for vapour pressures and gas mixtures, as Sections 1.4 and 1.5 will show.

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

(1.7)

(vi) Molarity (M). The molarity of a solution is the number of moles of solute in one litre (one cubic decimetre) of the solution. A '0.10 M HCl' bottle in the laboratory holds $0.10 m o l$ of hydrogen chloride in every litre. Molarity is convenient because volumes are easy to measure with pipettes and burettes.

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

(1.8)

where

$V_{solution}$: volume of the solution in litres

(vii) Molality (m). The molality is the number of moles of solute per kilogram of the solvent (not of the solution). A $0.50 m o l k g^{- 1}$ (0.50 m) sugar solution is made by dissolving $0.50 m o l$ of sugar in $1 k g$ of water, whatever volume results.

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

(1.9)

where

$w_{solvent}$: mass of the solvent in kilograms

Which measure should you trust when the temperature changes? Masses and mole numbers do not change on warming, but volumes do, because liquids expand. So mass percentage, ppm by mass, mole fraction and molality stay fixed, while molarity (and any volume-based measure) falls slightly as a solution warms. That is why the laws of Section 1.6, which are used over a range of temperatures, are written with molality.

Compare

Does the measure change when the solution is warmed?

Independent of temperature

mass percentage (w/w)
ppm by mass
mole fraction
molality

Changes with temperature

molarity
volume percentage
mass by volume percentage
anything per litre of solution

Example 1.1NCERT Example 1.1 · p. 4

Mole fraction from a mass percentage

A car's cooling liquid contains 40 % propylene glycol ( $C_{3} H_{8} O_{2}$ ) by mass in water. Find the mole fraction of each component. Use $M (C_{3} H_{8} O_{2}) = 76 g m o l^{- 1}$ and $M (H_{2} O) = 18 g m o l^{- 1}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 1.2NCERT Example 1.2 · p. 4

Molarity from a mass and a volume

A student dissolves $2.65 g$ of anhydrous sodium carbonate ( $N a_{2} C O_{3}$ , $M = 106 g m o l^{- 1}$ ) in water and makes the volume up to $200 m L$ in a volumetric flask. What is the molarity?

Try it yourself first. Then reveal the solution one step at a time.

Example 1.3NCERT Example 1.3 · p. 5

Molality in a non-aqueous solvent

What is the molality of a solution made by dissolving $1.28 g$ of naphthalene ( $C_{10} H_{8}$ ) in $40.0 g$ of toluene?

Try it yourself first. Then reveal the solution one step at a time.

Example

Converting with the help of density

A 12.0 % (w/w) aqueous solution of sodium chloride ( $M = 58.5 g m o l^{- 1}$ ) has a density of $1.086 g m L^{- 1}$ . Find its molality and its molarity.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

A solution is labelled 0.50 m and another 0.50 M. Which statement is true?

The 0.50 m label refers to a kilogram of solvent; the 0.50 M label to a litre of solutionBoth contain 0.50 mol of solute per kilogram of waterThey are the same, since a litre of water weighs a kilogramThe 0.50 M label refers to a kilogram of solution

Show answer

Answer (A) The 0.50 m label refers to a kilogram of solvent; the 0.50 M label to a litre of solution

Lower-case m is molality (per kg of solvent); capital M is molarity (per litre of solution). They are close only for dilute aqueous solutions, where a litre of solution holds nearly a kilogram of water.

Practice · 5 questions

Practice: concentration (Intext Questions 1.1–1.5): questions with answers

NumericalEasyNCERT Intext Question 1.1
$18.0 g$ of toluene is dissolved in $132 g$ of hexane. Find the mass percentage of each component.
Show solution
AnswerToluene 12.0 %, hexane 88.0 %.
Total mass $= 18.0 + 132 = 150 g$ . Toluene: $18.0 / 150 \times 100 = 12.0 %$ . Hexane: $132 / 150 \times 100 = 88.0 %$ (or $100 - 12.0$ ).
NumericalMediumNCERT Intext Question 1.2
A solution of acetone ( $C_{3} H_{6} O$ , $58 g m o l^{- 1}$ ) in chloroform ( $C H C l_{3}$ , $119.5 g m o l^{- 1}$ ) is 25.0 % acetone by mass. Find the mole fraction of acetone.
Show solution
In $100 g$ : acetone $25.0 / 58 = 0.431 m o l$ ; chloroform $75.0 / 119.5 = 0.628 m o l$ . So $x_{acetone} = 0.431 / 1.059 = 0.407$ and $x_{chloroform} = 0.593$ .
3
NumericalMediumNCERT Intext Question 1.3
Find the molarity of each solution.
1. $25.0 g$ of blue vitriol, $C u S O_{4} \cdot 5 H_{2} O$ ( $249.7 g m o l^{- 1}$ ), dissolved to make $2.00 L$ of solution.
  M
  Show solution
  $n = 25.0 / 249.7 = 0.1001 m o l$ , so molarity $= 0.1001 / 2.00 = 0.0501 M$ . The water of crystallisation simply joins the solvent.
2. $20.0 m L$ of $1.5 M$ nitric acid diluted with water to $250 m L$ .
  M
  Show solution
  Moles of acid $= 1.5 \times 0.0200 = 0.030 m o l$ ; new molarity $= 0.030 / 0.250 = 0.12 M$ . (Equivalently $M_{1} V_{1} = M_{2} V_{2}$ .)
NumericalMediumNCERT Intext Question 1.4
What mass of glucose ( $180 g m o l^{- 1}$ ) is needed to make $1.50 k g$ of a $0.200 m o l k g^{- 1}$ aqueous glucose solution?
g
Show solution
$1 k g$ of water needs $0.200 m o l = 36.0 g$ of glucose, making $1036 g$ of solution. Scaling to $1500 g$ of solution: glucose $= 36.0 \times 1500 / 1036 = 52.1 g$ (with $1448 g$ of water).
5
NumericalHardNCERT Intext Question 1.5
An aqueous solution of sodium nitrate ( $N a N O_{3}$ , $85.0 g m o l^{- 1}$ ) is 15.0 % by mass and has a density of $1.107 g m L^{- 1}$ . Work out three measures of its concentration.
1. Molality
  mol/kg
  Show solution
  $n = 15.0 / 85.0 = 0.1765 m o l$ in $0.0850 k g$ of water: $m = 2.08 m o l k g^{- 1}$ .
2. Molarity
  M
  Show solution
  $V = 100 / 1.107 = 90.3 m L$ , so molarity $= 0.1765 / 0.0903 = 1.95 M$ .
3. Mole fraction of $N a N O_{3}$
  Show solution
  Water: $85.0 / 18 = 4.722 m o l$ . $x = 0.1765 / (0.1765 + 4.722) = 0.0360$ .

1.3 Solubility

Keep adding sugar to a glass of cold water and a point comes when crystals lie undissolved at the bottom however long you stir. The solubility of a substance is the largest amount of it that will dissolve in a stated amount of solvent at a given temperature (and, for gases, pressure). It depends on what the solute and solvent are, and on the temperature and pressure. We look first at solids in liquids, then at gases.

Definition

Solubility

The largest quantity of a substance that a given quantity of solvent can hold in solution at a stated temperature and pressure.

ExampleAbout $36 g$ of common salt dissolves in $100 g$ of water at room temperature.

1.3.1 Solubility of a Solid in a Liquid

Why does turmeric colour a ghee tadka so strongly but tint water only faintly? Curcumin, its yellow pigment, is a largely non-polar molecule, and it mixes far better with the non-polar fat than with water. Common salt does the opposite: it dissolves readily in water and not at all in coconut oil. Petrol and kerosene lift grease stains that water cannot touch. The rule of thumb is 'like dissolves like': a solute dissolves well in a solvent whose intermolecular forces resemble its own, polar in polar and non-polar in non-polar.

Watch what happens at the molecular level when a solid sits in a solvent. Particles leave the crystal surface and spread into the liquid; this is dissolution. At the same time, dissolved particles wandering near the crystal stick back onto it; this is crystallisation. At first dissolution wins, but as the solution fills up, crystallisation speeds up, until the two rates are equal (Fig. 1.3). From then on the amount dissolved stays constant although both processes continue: a dynamic equilibrium.

Dissolution equilibrium

solute + solvent ⇌ solution

(1.10)

3/3The two rates are equal. The amount dissolved no longer changes, though both processes go on.

Fig. 1.3Dissolution and crystallisation run at the same time; a saturated solution is where their rates match.NCERT p. 6

A beaker on the left holds a pile of crystals at the bottom and dots for dissolved particles; two bars on the right compare the rate of dissolving with the rate of crystallising. In step 1 the pile is large, there are few dissolved particles, the dissolving bar is long and the crystallising bar is very short. In step 2 the pile is smaller, there are many more dissolved particles, and the crystallising bar has grown while the dissolving bar is unchanged. In step 3 some crystals still remain, the dissolved particles have stopped increasing, and the two bars are equal in length: the solution is saturated.

A solution in this state of equilibrium with undissolved solute is a saturated solution (संतृप्त विलयन): at that temperature and pressure it holds as much solute as it can, so its concentration is the solubility. A solution that can still dissolve more solute is unsaturated.

Definition

Saturated solution

A solution in dynamic equilibrium with undissolved solute; it holds the maximum amount of solute possible at that temperature and pressure.

Effect of temperature. Because equation (1.10) is an equilibrium, Le Chatelier's principle predicts how it responds to heating. If dissolving a nearly saturated solid absorbs heat ( $Δ_{sol} H > 0$ , endothermic), raising the temperature pushes the equilibrium to the right and the solubility increases. If dissolving releases heat ( $Δ_{sol} H < 0$ , exothermic), the solubility falls on heating. Both behaviours are seen in practice. Potassium nitrate dissolves endothermically and its solubility rises steeply, while for common salt the heat effect is tiny and so is the change (Table 1.2).

Table 1.2Approximate solubility in grams per 100 g of water

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

Effect of pressure. Squeezing hardly changes the volume of a solid or a liquid, so pressure has no noticeable effect on how much solid dissolves in a liquid. Gases are different, as we see next.

1.3.2 Solubility of a Gas in a Liquid

Gases differ enormously in how well they dissolve in water. Nitrogen barely dissolves, and oxygen only a little, yet that small amount of dissolved oxygen keeps every fish and water plant alive. Ammonia, at the other extreme, dissolves so readily that one volume of cold water takes up several hundred volumes of the gas. For any one gas, though, two conditions control the amount dissolved: pressure and temperature.

Effect of pressure. Think of the boundary between gas and liquid as a busy doorway. The rate at which gas molecules go in depends on how many arrive at the surface each second, which is proportional to the number of gas molecules per unit volume, that is, to the gas pressure. The rate at which they come out depends on how many are already dissolved. At equilibrium the two rates match, so the amount dissolved must rise in step with the pressure. Push a piston down on the gas (Fig. 1.4) and for a moment more molecules enter than leave; the dissolved amount grows until the outward traffic catches up, and the system settles at a higher pressure with more gas in solution.

3/3More gas dissolves until the point is back on the line, at a higher pressure and higher mole fraction.

Fig. 1.4Compressing the gas raises its pressure; more gas then dissolves until the system returns to the Henry's-law line.NCERT Fig. 1.1 · p. 7

On the left a closed cylinder holds a liquid with a gas above it and a piston at the top; on the right a graph plots the partial pressure of the gas against its mole fraction in the liquid, with a straight line through the origin. In step 1 the piston is high, the gas is sparse, few gas particles are dissolved, and a point sits on the line at low pressure. In step 2 the piston has been pushed down: the gas is crowded but the dissolved amount has not yet changed, so the point jumps straight up, above the line. In step 3 more gas has dissolved and the point has moved back onto the line at a higher pressure and a larger mole fraction.

LawHenry's law

At a fixed temperature, the partial pressure of a gas in equilibrium with a solution is directly proportional to the mole fraction of that gas in the solution.

Henry's law

p = K_{H} x

(1.11)

where

$p$: partial pressure of the gas above the solution
$x$: mole fraction of the gas in the solution
$K_{H}$: Henry's law constant, in pressure units (it depends on the gas, the solvent and the temperature)

The constant $K_{H}$ is a property of the gas–solvent pair. Rearranging, $x = p / K_{H}$ : at a given pressure, a gas with a larger $K_{H}$ dissolves less. Table 1.3 shows helium and nitrogen at the insoluble end and carbon dioxide, which reacts slightly with water, at the soluble end. The oxygen values also show $K_{H}$ rising with temperature, which means gases dissolve less in warm water.

Table 1.3Approximate Henry's law constants for some gases in waterNCERT 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

How much oxygen does a pond hold?

Air in contact with a pond at $298 K$ supplies oxygen at a partial pressure of $0.21 b a r$ . Taking $K_{H} (O_{2}) = 44 k b a r$ , find the amount of dissolved oxygen in $1.00 L$ of the water, in millimoles and in milligrams. Take one litre of water as $55.5 m o l$ of $H_{2} O$ .

Try it yourself first. Then reveal the solution one step at a time.

Henry's law turns up in industry, in medicine and in the mountains:

Fizzy drinks. Soda water and soft drinks are bottled under a high pressure of carbon dioxide so that a lot of the gas dissolves. Open the cap and the pressure above the drink falls to that of the CO2 in air, which is tiny, so the extra gas leaves as a stream of bubbles.
Divers and the bends. A diver 30 m under the sea off the Andamans breathes air at about four times atmospheric pressure, so about four times as much nitrogen dissolves in the blood and tissues. If the diver surfaces too fast, the nitrogen comes out of solution as bubbles in the blood vessels and joints, a painful and dangerous condition called the bends. Deep divers often breathe mixtures in which helium, which is even less soluble than nitrogen, replaces part of the nitrogen.
High altitudes. At Leh, about 3500 m up, the air pressure and so the partial pressure of oxygen are roughly two-thirds of their sea-level values. Less oxygen dissolves in the blood, and newcomers can feel weak, breathless and muddled, a condition called anoxia; visitors are told to rest for a day or two to acclimatise.

Effect of temperature. Heat a pan of water and tiny bubbles cling to its sides long before it boils: dissolved air is coming out. A dissolved gas molecule is like a gas molecule that has condensed into the liquid, and condensing releases heat, so dissolving a gas is exothermic. By Le Chatelier's principle, raising the temperature shifts the equilibrium back towards the free gas, and the solubility falls. This is why a warm pond holds less oxygen than a cold one, and why fish crowd near the surface on hot, still afternoons.

Try this

Warm soda, cold soda

You’ll need

two identical sealed bottles of soda
a refrigerator
a sink

Predict first

Which bottle will foam more when opened?

The cold oneThe warm oneBoth the same, since they hold the same gas

Take two identical unopened bottles of the same soda. Keep one in the fridge for a few hours and leave the other in a warm room.
Hold each bottle over a sink, pointed away from faces, and open them one after the other.
Compare how much fizz rises and how quickly. Then leave both open for ten minutes and taste a sip of each.

Table 1.4What did you see?

Bottle	Fizz on opening	After 10 min
Cold
Warm

Your entries are saved automatically.

Practice · 2 questions

Practice: Henry's law (Intext Questions 1.6–1.7): questions with answers

NumericalMediumNCERT Intext Question 1.6
At $273 K$ and a carbon dioxide pressure of $1.00 b a r$ , the solubility of $C O_{2}$ in water is $0.0767 m o l k g^{- 1}$ . Calculate Henry's law constant in bar and in pascals. Compare it with the value at $298 K$ in Table 1.3.
bar
Show solution
Moles of water in $1 k g$ : $1000 / 18.02 = 55.5 m o l$ . Mole fraction of CO2: $x = 0.0767 / (0.0767 + 55.5) = 1.38 \times 10^{- 3}$ .
$K_{H} = p / x = 1.00 b a r / 1.38 \times 10^{- 3} = 725 b a r = 7.25 \times 10^{7} P a$ . At $298 K$ the constant is about $1.7 k b a r$ , more than twice as large, so cold water dissolves more than twice as much CO2: one reason a fizzy drink tastes flatter when warm.
NumericalMediumNCERT Intext Question 1.7
A $750 m L$ bottle of soda is sealed at $298 K$ under a carbon dioxide pressure of $3.0 b a r$ . Taking $K_{H} (C O_{2}) = 1.66 k b a r$ and treating the drink as water, how much CO2 (in grams) is dissolved?
g
Show solution
$x = p / K_{H} = 3.0 / 1660 = 1.81 \times 10^{- 3}$ . Moles of water: $750 / 18.02 = 41.6 m o l$ .
$n_{C O_{2}} \approx x \times n_{water} = 1.81 \times 10^{- 3} \times 41.6 = 0.075 m o l$ , and $0.075 \times 44 = 3.3 g$ of CO2.

1.4 Vapour Pressure of Liquid Solutions

Section 1.3.2 dealt with gases dissolved in liquids. Two kinds of liquid solution remain, and this section treats both: mixtures of two liquids, such as the alcohols in a hand sanitiser, where each component can evaporate, and solutions of a solid in a liquid, such as sugar syrup, where normally only the solvent can. In both cases the key measurable property is the vapour pressure of the solution.

Recall what vapour pressure is. Pour a liquid into a closed flask and some molecules escape from its surface into the space above. Those vapour molecules also strike the surface and some are recaptured. The vapour fills up until escape and capture happen at the same rate. The pressure the vapour then exerts is the liquid's vapour pressure. It depends only on the liquid and on the temperature, not on how much liquid there is, and it rises steeply as the liquid gets warmer. Try it in Fig. 1.5.

SimulationVapour above a liquid in a closed container

A container of water particles with a lid, a thermometer and a temperature slider from 20 to 100 degrees Celsius, plus a button to take the lid off or put it on. With the lid on, particles leave the liquid surface and wander in the space above; some of them fall back. After a while the number above the liquid stops growing. At a higher temperature more particles are in the vapour at any moment. With the lid off, escaped particles drift away and the liquid slowly disappears.

Fig. 1.5Water in a lidded container. Press Play, then change the temperature or lift the lid.

Try this

With the lid on at 40 °C, press Play and watch the space above the water. Does the number of vapour particles keep growing, or does it settle?
Raise the temperature to 80 °C. Once it settles again, are there more or fewer particles in the vapour than before? What does that say about the vapour pressure?
Take the lid off. Why can a balance between escape and capture never be reached now?

1.4.1 Vapour Pressure of Liquid-Liquid Solutions

Sniff a nail-polish remover made from two solvents and you smell both: both are in the vapour. Seal such a mixture of two volatile liquids, 1 and 2, in a container and wait. Each liquid sends molecules into the space above until evaporation and condensation balance, and the vapour then exerts a total pressure $p_{total}$ made up of two partial vapour pressures, $p_{1}$ and $p_{2}$ . Our question is how these depend on the liquid's composition, measured by the mole fractions $x_{1}$ and $x_{2}$ .

The French chemist François-Marie Raoult answered this in the 1880s by measuring the vapour pressures of many mixtures. His result, Raoult's law, is strikingly simple: each component evaporates as if it were pure, scaled down by its share of the molecules in the liquid.

LawRaoult's law (for volatile components)

In a solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction in the liquid; the constant of proportionality is the vapour pressure of that component when pure, at the same temperature.

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

(1.12)

where

$p_{1}^{\circ}$: vapour pressure that liquid 1 would have on its own, at the temperature of the mixture

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

(1.13)

Dalton's law of partial pressures says the total vapour pressure is the sum of the partial pressures. Substituting Raoult's law, and then $x_{1} = 1 - x_{2}$ , gives the total pressure in terms of one mole fraction only:

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) is a straight line, and reading it carefully gives three conclusions. First, the total vapour pressure can be written using the mole fraction of either component alone. Second, it changes linearly with $x_{2}$ , from $p_{1}^{\circ}$ for pure liquid 1 to $p_{2}^{\circ}$ for pure liquid 2. Third, whether adding more of component 1 raises or lowers the total depends on which pure liquid is more volatile: if $p_{1}^{\circ} < p_{2}^{\circ}$ , the total falls as $x_{1}$ grows. Fig. 1.6 builds the line from its two parts.

4/4The tops of the stacks lie on a straight line from $p_{1}^{\circ}$ to $p_{2}^{\circ}$ : equation (1.16).

Fig. 1.6For an ideal mixture each partial pressure is a straight line, so their sum, the total vapour pressure, is a straight line too.NCERT Fig. 1.3 · p. 10

A graph with mole fraction of component 2 along the bottom from 0 to 1 and vapour pressure upwards. Pure component 1 has the lower vapour pressure, on the left axis; pure component 2 has the higher one, on the right axis. Step 1 draws a dashed line for the partial pressure of component 1, falling from its pure value on the left to zero on the right. Step 2 adds a dashed line for component 2, rising from zero on the left to its pure value on the right. Step 3 stacks the two partial pressures as bars at mole fractions 0.25, 0.5 and 0.75, one segment on top of the other. Step 4 joins the tops of the bars with a solid straight line, the total vapour pressure, which runs from the pure value of component 1 to that of component 2.

What is in the vapour? By Dalton's law each component's partial pressure is its mole fraction in the vapour, written $y$ , times the total pressure. So $y_{i} = p_{i} / p_{total}$ :

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

(1.17)

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

(1.18)

Composition of the vapour

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

(1.19)

Because the more volatile liquid contributes more than its share of the pressure, the vapour is always richer in the more volatile component than the liquid it came from. Condense that vapour and evaporate it again, and it becomes richer still; this is the idea behind fractional distillation. Explore it in Fig. 1.7: with the deviation slider in the middle, set the two pure vapour pressures, move the composition, and compare $y_{2}$ with $x_{2}$ .

InteractiveRaoult's law explorer

An interactive graph of vapour pressure against the mole fraction of component 2. Sliders set the vapour pressures of the two pure liquids (component 2 can be made non-volatile by setting it to zero), the mole fraction of component 2 in the liquid, and how strongly unlike molecules attract each other compared with like ones. The graph shows the partial pressure lines of both components, the total vapour pressure and, when the mixture is non-ideal, the straight Raoult's-law line for comparison. Readouts give the two partial pressures, the total, and the mole fraction of component 2 in the vapour, with a message saying which component the vapour is richer in, or that the mixture is at an azeotrope. A preset loads the methanol and ethanol mixture of Example 1.5.

Fig. 1.7Set the pure vapour pressures and the liquid composition; read the partial and total pressures and the vapour composition. The deviation slider makes the mixture non-ideal (Section 1.5).

Example 1.5NCERT Example 1.5 · p. 11

Total vapour pressure and vapour composition

Methanol and ethanol form a nearly ideal solution. At $313 K$ their pure vapour pressures are $35.4 k P a$ and $17.8 k P a$ . A mixture is made from $32.0 g$ of methanol ( $C H_{3} O H$ , $32 g m o l^{- 1}$ ) and $69.0 g$ of ethanol ( $C_{2} H_{5} O H$ , $46 g m o l^{- 1}$ ). Find (i) the vapour pressure of the mixture at $313 K$ and (ii) the mole fraction of each alcohol in the vapour.

Try it yourself first. Then reveal the solution one step at a time.

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

Write Raoult's law, $p_{i} = p_{i}^{\circ} x_{i}$ , under Henry's law, $p = K_{H} x$ , and they look like one law with different constants: in each, a component's partial pressure above the solution is proportional to its mole fraction in it. That is no coincidence. A dissolved gas is simply a component volatile enough to be a gas at the working temperature, and Henry's constant plays the part of its vapour pressure. So Raoult's law can be seen as Henry's law in the particular case $K_{H} = p_{i}^{\circ}$ .

Compare

Two laws, one shape

Henry's law

$p = K_{H} x$
used for a gas dissolved in a liquid
$K_{H}$ found by experiment for each gas–solvent pair

Raoult's law

$p_{i} = p_{i}^{\circ} x_{i}$
used for a volatile liquid in a solution
the constant is the vapour pressure of the pure liquid

1.4.3 Vapour Pressure of Solutions of Solids in Liquids

Many everyday solutions are a solid dissolved in a liquid: salt in pickling brine, sugar in a syrup, iodine in alcohol as a tincture, camphor in coconut oil. Usually the solid is non-volatile: its particles practically never escape into the vapour. Then the whole vapour pressure of the solution comes from the solvent, and experiments show it is always lower than that of the pure solvent at the same temperature.

Picture the liquid surface as a row of exits through which molecules can leave for the vapour (Fig. 1.8). In the pure solvent every exit is in use. Dissolve a non-volatile solute and some surface places are taken by solute particles that never leave, so fewer solvent molecules escape per second. Vapour molecules still return to the liquid at the old rate for a given vapour density, so escape and return balance at a smaller vapour density, that is, at a lower vapour pressure. (A deeper treatment shows that the solvent's escaping tendency is proportional to its mole fraction; the surface picture is a convenient way to imagine it.)

3/3Capture matches the slower escape at a smaller vapour density: a lower vapour pressure.

Fig. 1.8A non-volatile solute takes up part of the surface, so fewer solvent molecules escape and the vapour pressure is lower.NCERT Fig. 1.4 · p. 12

A magnified strip of liquid surface seen from the side, with vapour above it. In step 1, pure solvent: every surface site is a round solvent molecule, five arrows show molecules escaping, and the vapour above is fairly crowded. In step 2 solute particles, drawn as squares, occupy three of the ten surface sites; they have no escape arrows, and only three solvent molecules are escaping. In step 3 the vapour above the solution has settled with fewer molecules than above the pure solvent, labelled lower vapour pressure.

How much lower? The fall depends on how much solute is present, not on what it is: half a mole of glucose and half a mole of glycerol, each dissolved in a kilogram of water, lower its vapour pressure by almost exactly the same amount. The general form of Raoult's law covers this case: in any solution, every volatile component contributes a partial vapour pressure in proportion to its mole fraction. With the solvent as component 1 and a non-volatile solute as component 2, only component 1 contributes:

Solution of a non-volatile solute

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

(1.20)

A graph of $p_{1}$ against $x_{1}$ is a straight line from zero (no solvent) to $p_{1}^{\circ}$ (pure solvent). You can see it in the explorer above by setting the vapour pressure of component 2 to zero.

1.5 Ideal and Non-ideal Solutions

Raoult's law fits some liquid mixtures perfectly and others badly. That gives a natural way to sort liquid–liquid solutions into two families: those that obey it at every composition, and those that do not.

1.5.1 Ideal Solutions

An ideal solution obeys Raoult's law over the whole range of composition. Two further properties always go with this. Mixing the pure components neither releases nor absorbs heat, and the volume of the mixture is exactly the sum of the volumes mixed:

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

(1.21)

Definition

Ideal solution

A solution that obeys Raoult's law at every composition, with $Δ_{mix} H = 0$ and $Δ_{mix} V = 0$ .

The molecular reason is easiest to see with two components A and B. In the pure liquids, each molecule is held only by its own kind: A–A or B–B attractions. In the mixture, A–B attractions appear as well. If an A–B attraction is about as strong as the A–A and B–B attractions it replaces, a molecule cannot 'tell' whether its neighbours are A or B. Its escaping tendency is unchanged, so Raoult's law holds, no energy is gained or lost on mixing, and the molecules pack as before. Truly ideal solutions are rare, but mixtures of very similar molecules come close: n-pentane with n-hexane, chlorobenzene with bromobenzene, and the methanol–ethanol pair of Example 1.5.

1.5.2 Non-ideal Solutions

A non-ideal solution fails to follow Raoult's law across the full range of composition: its vapour pressure is higher or lower than equation (1.16) predicts. If it is higher, the solution shows a positive deviation from Raoult's law; if lower, a negative deviation (Fig. 1.9). Once again, the cause is the balance of intermolecular forces.

3/3A–B attractions stronger: molecules are held back and the curve sags below the line.

Fig. 1.9Weaker attractions between unlike molecules push the vapour pressure above Raoult's line; stronger ones pull it below.NCERT Fig. 1.6 · p. 13

A graph of total vapour pressure against the mole fraction of component 2, with the pure-component vapour pressures on the left and right axes. Step 1 shows the straight dashed Raoult's-law line joining them, labelled ideal, with a note that unlike attractions are as strong as like ones. Step 2 adds a curve bowing above the line, labelled positive deviation, with a note that unlike molecules attract each other more weakly. Step 3 adds a curve sagging below the line, labelled negative deviation, with a note that unlike molecules attract each other more strongly. The ideal line stays visible throughout for comparison.

Positive deviation. Here the A–B attractions are weaker than the A–A and B–B attractions. Each kind of molecule is held less firmly in the mixture than in its pure liquid, finds it easier to escape, and the vapour pressure rises above the Raoult's-law value. Ethanol mixed with hexane is a clear case. In pure ethanol the molecules are linked by hydrogen bonds; hexane molecules slip between them and break many of those bonds while offering only weak attractions in return. Breaking the hydrogen bonds costs energy, so mixing absorbs heat ( $Δ_{mix} H > 0$ ), and the looser packing usually makes the volume grow ( $Δ_{mix} V > 0$ ).

Negative deviation. Here the A–B attractions are stronger than the like attractions. Each molecule is held more tightly in the mixture, escapes less readily, and the vapour pressure falls below the Raoult's-law value. Chloroform with diethyl ether is an example: the hydrogen of $C H C l_{3}$ , made slightly positive by its three chlorine neighbours, forms a hydrogen bond to the oxygen of the ether, an attraction neither pure liquid has. Forming these new attractions releases energy, so the mixture warms ( $Δ_{mix} H < 0$ ) and usually contracts a little ( $Δ_{mix} V < 0$ ). Hydrochloric acid and water, where strong ion–water attractions form, deviate the same way.

Compare

Ideal and non-ideal liquid mixtures

Ideal

A–B ≈ A–A, B–B
obeys Raoult's law
$Δ_{mix} H = 0$ , $Δ_{mix} V = 0$
e.g. n-pentane + n-hexane

Positive deviation

A–B weaker
vapour pressure above Raoult
$Δ_{mix} H > 0$ , $Δ_{mix} V > 0$
e.g. ethanol + hexane
may form a minimum-boiling azeotrope

Negative deviation

A–B stronger
vapour pressure below Raoult
$Δ_{mix} H < 0$ , $Δ_{mix} V < 0$
e.g. chloroform + diethyl ether
may form a maximum-boiling azeotrope

Azeotropes. When the deviation is large, the total vapour pressure curve develops a maximum or a minimum. At that special composition the liquid and the vapour it gives off have exactly the same composition, so boiling does not change the mixture and the liquid boils at a constant temperature, just like a pure substance. Such a mixture is an azeotrope (स्थिरक्वाथी), and no amount of fractional distillation can separate it further.

A large positive deviation gives a vapour-pressure maximum, which means the mixture boils below both pure liquids: a minimum-boiling azeotrope. Ethanol and water form one at about 95.6 % ethanol by mass, boiling at about 351 K, so distilling a fermented liquid can never give pure ethanol.
A large negative deviation gives a vapour-pressure minimum and a maximum-boiling azeotrope, boiling above both pure liquids. Hydrochloric acid and water form one at about 20 % HCl by mass, boiling at about 382 K (109 °C), which is why concentrated hydrochloric acid cannot be made by simply boiling off water.

Definition

Azeotrope

A liquid mixture of fixed composition that boils at a constant temperature and gives vapour of the same composition, so fractional distillation cannot separate it.

Try both kinds of deviation in Fig. 1.7: slide the attraction control towards 'weaker' or 'stronger' and watch for the point where the vapour and the liquid compositions become equal.

Quick check

1 of 3

Mixing two liquids warms the mixture. What kind of deviation from Raoult's law do you expect?

Positive, because the mixture is warmer and evaporates moreNone, because warming does not affect vapour pressurePositive, because the molecules have more energyNegative, because heat release means new, stronger A–B attractions

Show answer

Answer (D) Negative, because heat release means new, stronger A–B attractions

A release of heat (

Δ_{mix} H < 0

) shows that the A–B attractions formed are stronger than the A–A and B–B ones they replaced. Stronger holding means a lower escaping tendency: a negative deviation. Comparing at the same temperature removes the warming effect.

Practice · 1 question

Practice: liquid mixtures (Intext Question 1.8): questions with answers

1
NumericalMediumNCERT Intext Question 1.8
Pure liquids P and Q have vapour pressures of $32.0 k P a$ and $84.0 k P a$ at $330 K$ , and form an ideal solution. A mixture of them has a total vapour pressure of $60.0 k P a$ at this temperature. Find the composition of (a) the liquid and (b) the vapour.
1. Mole fraction of Q in the liquid
  Show solution
  $60.0 = 32.0 + (84.0 - 32.0) x_{Q}$ , so $x_{Q} = 28.0 / 52.0 = 0.538$ and $x_{P} = 0.462$ .
2. Mole fraction of Q in the vapour
  Show solution
  $p_{Q} = 0.538 \times 84.0 = 45.2 k P a$ , so $y_{Q} = 45.2 / 60.0 = 0.754$ and $y_{P} = 0.246$ . The vapour is richer in the more volatile Q.

1.6 Colligative Properties and Determination of Molar Mass

A non-volatile solute does more than lower the solvent's vapour pressure (Section 1.4.3). The same change makes the solution boil at a higher temperature and freeze at a lower one, and it makes solvent push its way into the solution through a suitable membrane. That gives four measurable effects:

the relative lowering of the solvent's vapour pressure,
the elevation of its boiling point,
the depression of its freezing point, and
the osmotic pressure of the solution.

Here is the remarkable part. In a dilute solution each effect is set by the count of solute particles compared with solvent particles, whatever those particles are: $0.1 m o l$ of glucose and $0.1 m o l$ of urea in the same water give the same four effects, even though a glucose molecule is three times as heavy. Such effects are called colligative properties (अणुसंख्य गुणधर्म), from the Latin colligatus, 'bound together', because they depend on the particles taken as a collection. Since they count particles, measuring one of them for a known mass of solute tells us how heavy each mole of particles is: the molar mass.

Definition

Colligative property

A property of a dilute solution that depends on the number of solute particles relative to solvent particles, and not on their chemical identity.

1.6.1 Relative Lowering of Vapour Pressure

Raoult showed that the lowering of a solvent's vapour pressure depends only on the concentration of solute particles. Starting from equation (1.20), a short derivation turns this into a tool for finding molar masses.

DerivationFrom Raoult's law to a molar mass

For a solvent (1) containing a non-volatile solute (2), Raoult's law gives the solvent's vapour pressure:

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

(1.22)

The lowering of the vapour pressure is the difference from the pure solvent:

\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)

In a binary solution $1 - x_{1} = x_{2}$ , so the lowering is proportional to the solute's mole fraction:

Δ p_{1} = x_{2} p_{1}^{\circ}

(1.24)

Dividing by $p_{1}^{\circ}$ gives the relative lowering of vapour pressure, a pure number equal to the mole fraction of the solute. (With several non-volatile solutes, $x_{2}$ becomes the sum of their mole fractions.)

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

(1.25)

Write $x_{2}$ with $n_{1}$ moles of solvent and $n_{2}$ moles of solute:

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

(1.26)

In a dilute solution $n_{2} ≪ n_{1}$ , so $n_{2}$ can be dropped from the denominator. Then, with masses $w_{1}$ , $w_{2}$ and molar masses $M_{1}$ , $M_{2}$ of solvent and solute, $n = w / M$ :

\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)

Everything in equation (1.28) except $M_{2}$ can be measured, so it gives the molar mass of the solute.

Example 1.6NCERT Example 1.6 · p. 16

Molar mass from a vapour-pressure measurement

Pure cyclohexane ( $C_{6} H_{12}$ , $84 g m o l^{- 1}$ ) has a vapour pressure of $13.00 k P a$ at $298 K$ . When $2.40 g$ of a non-volatile, non-electrolyte solid is dissolved in $60.0 g$ of cyclohexane, the vapour pressure falls to $12.66 k P a$ . Find the molar mass of the solid.

Try it yourself first. Then reveal the solution one step at a time.

1.6.2 Elevation of Boiling Point

Rice water with a pinch of salt boils a few hundredths of a degree later than plain water. Why? A liquid boils when its vapour pressure has risen to match the pressure of the air above it; for pure water that happens at $373.15 K$ , where the vapour pressure reaches $1.013 b a r$ . Dissolving a non-volatile solute lowers the vapour pressure at every temperature, so at $373.15 K$ the solution is still short of $1.013 b a r$ and has to be heated further before it boils. A solution therefore always boils above its pure solvent, as the first three steps of Fig. 1.10 show.

5/5The solution's curve meets the solid's at a lower temperature $T_{f}$ : the depression $Δ T_{f} = T_{f}^{\circ} - T_{f}$ .

Fig. 1.10Lowering the vapour-pressure curve moves its meeting point with the atmosphere line to a higher temperature, and its meeting point with the solid's curve to a lower one.NCERT Fig. 1.7 · p. 17

A schematic graph of vapour pressure upwards against temperature to the right. Step 1 draws a rising curve for the pure liquid solvent and a horizontal dashed line at atmospheric pressure, 1.013 bar; where they cross is the normal boiling point. Step 2 adds a second curve just below the first for the solution, which has a lower vapour pressure at every temperature. Step 3 marks where the solution curve reaches the atmospheric line, at a higher temperature: the gap between the two crossing points is the boiling-point elevation. Step 4 switches attention to the low-temperature end and adds a steeper curve for the solid solvent; it meets the pure liquid curve at the normal freezing point. Step 5 marks where the solution curve meets the solid curve, at a lower temperature: the gap is the freezing-point depression.

Like the vapour-pressure lowering, the rise depends on the number of solute particles and not their kind. In dilute solutions, measurements show that the rise in boiling point, called the elevation of boiling point $Δ T_{b} = T_{b} - T_{b}^{\circ}$ , grows in proportion to the molality of the solute:

Δ T_{b} \propto m

(1.29)

Boiling-point elevation

Δ T_{b} = K_{b} m

(1.30)

where

$K_{b}$: boiling point elevation constant (molal elevation or ebullioscopic constant) of the solvent, in $K k g m o l^{- 1}$
$m$: molality of the solute

$K_{b}$ is the elevation a $1 m o l k g^{- 1}$ solution would show if it stayed ideal; it belongs to the solvent (Table 1.5). If $w_{2}$ grams of a solute of molar mass $M_{2}$ are dissolved in $w_{1}$ grams of solvent, the molality is

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

(1.31)

and substituting into equation (1.30) gives the elevation, or, turned round, the molar mass from a measured elevation:

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

Boiling point of a sugar solution

A sweet-maker dissolves $68.4 g$ of sucrose ( $C_{12} H_{22} O_{11}$ , $342 g m o l^{- 1}$ ) in $400 g$ of water. At what temperature will this syrup start to boil at $1.013 b a r$ ? Take $K_{b} = 0.52 K k g m o l^{- 1}$ for water.

Try it yourself first. Then reveal the solution one step at a time.

Example 1.8NCERT Example 1.8 · p. 18

How many atoms in a sulfur molecule?

$2.56 g$ of sulfur is dissolved in $100 g$ of carbon disulphide ( $K_{b} = 2.34 K k g m o l^{- 1}$ ). The boiling point rises by $0.234 K$ . Find the molar mass of sulfur in this solution and the number of atoms in each molecule (atomic mass of S = 32.06).

Try it yourself first. Then reveal the solution one step at a time.

1.6.3 Depression of Freezing Point

Now look at the low-temperature end of Fig. 1.10 (steps 4 and 5). The solid solvent has its own vapour-pressure curve, steeper than the liquid's. Where the two curves cross, solid and liquid have equal vapour pressures, so neither can grow at the other's expense by way of the vapour; that crossing is the freezing point. When a dilute solution freezes, the crystals that separate are pure solvent, so what matters is where the solution's curve meets the solid solvent's curve. The solution's curve is lower, so it reaches the solid's curve only at a lower temperature: the freezing point falls.

For dilute solutions the depression of freezing point, $Δ T_{f} = T_{f}^{\circ} - T_{f}$ , is proportional to the molality:

Freezing-point depression

Δ T_{f} = K_{f} m

(1.34)

where

$K_{f}$: freezing point depression constant (molal depression or cryoscopic constant) of the solvent, in $K k g m o l^{- 1}$

Substituting the molality from equation (1.31) gives the depression, and the molar mass from a measured depression:

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)

Both constants depend only on the solvent, and thermodynamics gives them from the solvent's molar mass $M_{1}$ (in $g m o l^{- 1}$ ), its freezing and boiling points $T_{f}$ and $T_{b}$ (in kelvin), its enthalpies of fusion and vaporisation, and the gas constant $R$ :

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 1.5Ebullioscopic and cryoscopic constants of some solvents (temperatures in K, constants in K kg mol⁻¹)NCERT Table 1.3 · p. 19

Solvent	Boils	$K_{b}$	Freezes	$K_{f}$
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

Where does 1.86 come from?

Water melts at $273.15 K$ with $Δ_{fus} H = 6.01 k J m o l^{- 1}$ , and $M_{1} = 18.02 g m o l^{- 1}$ . Use equation (1.37) to calculate $K_{f}$ for water.

Try it yourself first. Then reveal the solution one step at a time.

Example 1.9NCERT Example 1.9 · p. 20

Keeping water liquid below 0 °C

Glycerol ( $C_{3} H_{8} O_{3}$ , $92 g m o l^{- 1}$ ) is added to water to protect stored samples from freezing. $46.0 g$ of glycerol is dissolved in $400 g$ of water. Calculate (a) the depression of the freezing point and (b) the freezing point of the solution. $K_{f} (water) = 1.86 K k g m o l^{- 1}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 1.10NCERT Example 1.10 · p. 20

Molar mass by Rast's camphor method

$0.150 g$ of an organic compound is melted together with $10.0 g$ of camphor. The mixture's melting point is $3.47 K$ below that of pure camphor ( $K_{f} = 37.7 K k g m o l^{- 1}$ ). Find the molar mass of the compound.

Try it yourself first. Then reveal the solution one step at a time.

1.6.4 Osmosis and Osmotic Pressure

Soak a handful of shrivelled raisins in water overnight and by morning they are plump. Sprinkle salt on cucumber slices for raita and, in a few minutes, they sit in a pool of their own water. A bunch of limp spinach perks up in a bowl of cold water. Each case involves a membrane: the skin of a raisin, the walls of plant cells. Seen at high enough magnification, such a membrane is a mesh with gaps only a little wider than a water molecule. Water can slip through; larger solute particles, and ions wrapped in water molecules, mostly cannot. A barrier like this is called a semipermeable membrane. Living cell membranes behave this way, and so do artificial ones such as cellophane and the thin polymer films inside water purifiers.

Suppose a semipermeable membrane separates pure water from a solution, as in Fig. 1.11. Water molecules hit the membrane from both sides, but on the solution side some of the 'hits' are solute particles that cannot pass, so more water moves into the solution than out of it. This net flow of solvent through a semipermeable membrane, from pure solvent (or a dilute solution) into a more concentrated solution, is called osmosis. Remember the direction: the solvent moves, always towards the side with more solute particles.

3/3Pressing on the solution with exactly $Π$ prevents any net flow; the levels stay equal.

Fig. 1.11Solvent flows into the solution until a pressure difference stops it; the pressure needed to prevent any flow is the osmotic pressure.NCERT Fig. 1.9 · p. 20

A U-shaped tube with a semipermeable membrane, drawn dashed, across the bottom of the bend. The left arm holds pure water and the right arm a solution with solute particles. In step 1 the two levels are equal and arrows show more water crossing from left to right than back. In step 2 the right-hand level has risen and the left-hand level fallen; the difference in height is marked h, and the net flow has stopped because the extra liquid pushes back. In step 3 a piston presses on the right-hand arm with the osmotic pressure, keeping the two levels equal so that no net flow happens at all.

Pressing on the solution can stop the flow. The smallest extra pressure on the solution side that keeps any solvent from getting in is the osmotic pressure (परासरण दाब) of the solution, $Π$ . It too is colligative: it depends on the number of solute particles, not on their kind. For dilute solutions, experiments show that it is proportional to the molar concentration $C$ and to the absolute temperature, with the gas constant $R$ as the constant of proportionality:

Definition

Osmotic pressure

The extra pressure you must put on a solution so that no solvent flows into it across a semipermeable membrane; it measures how strongly the solution draws in solvent.

Osmotic pressure

Π = C R T

(1.39)

where

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

Writing $C = n_{2} / V$ for $n_{2}$ moles of solute in $V$ litres of solution, and $n_{2} = w_{2} / M_{2}$ , gives a route to the molar mass:

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)

Equation (1.42) is the preferred method for proteins, polymers and other large molecules. A few grams of such a substance contain very few moles, so the boiling-point and freezing-point changes would be far too small to measure, but the osmotic pressure is still a readable few millibar. The measurement is made at room temperature, which matters for proteins that would be damaged by heating, and it uses molarity, which is convenient for solutions made up in volumetric flasks.

Example 1.11NCERT Example 1.11 · p. 22

Molar mass of a protein

$0.350 g$ of a purified protein is dissolved in water to make $50.0 m L$ of solution. Its osmotic pressure at $298 K$ is $2.61 \times 10^{- 3} b a r$ . Find the molar mass of the protein ( $R = 0.0831 L b a r m o l^{- 1} K^{- 1}$ ). Then estimate the freezing-point depression of the same solution.

Try it yourself first. Then reveal the solution one step at a time.

Why is a hospital drip made with '0.9 % saline' and never with plain water? Because red blood cells are themselves tiny bags of solution wrapped in a semipermeable membrane (Fig. 1.12). A 0.9 % (mass/volume) solution of sodium chloride, called normal saline, has the same osmotic pressure as the fluid inside the cells, so water moves in and out equally and the cells keep their shape. Solutions with equal osmotic pressures at the same temperature are called isotonic. Put the cells into plain water, which has a lower osmotic pressure (a hypotonic solution), and water floods in until they swell and may burst. Put them into strong brine, which has a higher osmotic pressure (a hypertonic solution), and water leaves them, so they shrivel.

1/3In 0.9 % saline, water enters and leaves equally: the cell keeps its shape.

Fig. 1.12Water moves across a cell membrane towards the side with the higher osmotic pressure.NCERT p. 22

One red blood cell drawn large in the centre, surrounded by the solution it is placed in. In step 1 the solution is isotonic, 0.9 percent salt: the cell keeps its normal dimpled disc shape and small arrows show water moving in and out equally. In step 2 the solution is hypotonic, plain water: arrows point inwards and the cell has swollen into a round ball close to bursting. In step 3 the solution is hypertonic, strong salt water: arrows point outwards and the cell has shrunk into a smaller crinkled shape.

Osmosis is at work all around you. Water enters plant roots partly by osmosis, and the pressure it builds inside cells keeps leaves and stems firm; that is why limp spinach recovers in water. A diet very high in salt raises the osmotic pressure of body fluids, so the body holds back water and tissues can puff up, a swelling called edema. Pickles, salted fish and fruit preserves such as aam ka murabba keep for months because any bacterium or mould that lands on them is surrounded by a strongly hypertonic liquid, loses water by osmosis and cannot grow.

Try this

Raisins in water and in syrup

You’ll need

about 30 raisins
two glasses
sugar and water
a spoon
a plate

Predict first

What will happen to the raisins in plain water?

They will shrink furtherThey will swell upNothing, since raisins are already dry

Fill one glass with tap water and a second with a thick sugar syrup (about four spoons of sugar in half a glass of water, stirred until clear).
Count ten raisins of similar size into each glass. Keep ten more on a dry plate for comparison.
After eight to twelve hours, take the raisins out, pat them dry and compare their size and feel.

Table 1.6Raisins after soaking

Kept in	Size	Feel
Tap water
Sugar syrup
Dry plate

Your entries are saved automatically.

1.6.5 Reverse Osmosis and Water Purification

If the pressure on the solution side is raised above its osmotic pressure, the flow runs the other way: pure solvent is squeezed out of the solution, through the membrane, leaving the solute behind. This is reverse osmosis (RO), and it is the working principle of the RO purifier in many Indian kitchens as well as of large plants that turn seawater into drinking water (Fig. 1.13).

3/3Dissolved salts stay behind and leave, more concentrated, in the reject water.

Fig. 1.13In reverse osmosis, a pump pushes water through the membrane faster than osmosis would pull it back; salts stay behind in the reject stream.NCERT Fig. 1.11 · p. 23

A flow diagram of a home water purifier. Tap water enters from the left into a pump, which drives it at high pressure into a long membrane module drawn as a tube split lengthwise by a dashed membrane. In step 1 the feed water flows in and the pump is labelled pressure greater than the osmotic pressure. In step 2 water molecules cross the membrane into the lower channel and flow out to a storage tank labelled purified water. In step 3 the dissolved salts, drawn as small squares, stay above the membrane and leave through a separate outlet to the drain labelled reject water, more concentrated than the feed.

Making this work took a membrane tough enough for pressures of tens of bar. Early reverse-osmosis membranes were thin films of cellulose acetate; most purifiers and plants now use an even thinner polyamide layer cast on a porous backing that takes the strain. Either way, the film lets water molecules through while blocking dissolved ions and most other impurities. Many home purifiers send two or three litres of reject water to the drain for every litre they purify, so that water is worth collecting for washing or mopping. On a city scale, Chennai draws part of its drinking water from seawater reverse-osmosis plants at Minjur and Nemmeli.

Example

Minimum pressure to desalinate seawater

Treat seawater as $35 g$ of sodium chloride ( $58.5 g m o l^{- 1}$ ) per litre at $298 K$ . Each formula unit gives two ions in solution (Section 1.7). Estimate the smallest pressure a reverse-osmosis plant must apply. Use $R = 0.0831 L b a r m o l^{- 1} K^{- 1}$ .

Try it yourself first. Then reveal the solution one step at a time.

Practice · 4 questions

Practice: colligative properties (Intext Questions 1.9–1.12): questions with answers

NumericalMediumNCERT Intext Question 1.9
The vapour pressure of pure water at $303 K$ is $4.24 k P a$ . $30.0 g$ of glucose ( $180 g m o l^{- 1}$ ) is dissolved in $450 g$ of water. Find the vapour pressure of water over this solution and the relative lowering.
kPa
Show solution
$n_{2} = 30.0 / 180 = 0.1667 m o l$ ; $n_{1} = 450 / 18 = 25.0 m o l$ ; $x_{2} = 0.1667 / 25.17 = 0.00662$ .
Relative lowering $= x_{2} = 0.00662$ . Vapour pressure $p_{1} = 4.24 \times (1 - 0.00662) = 4.21 k P a$ .
NumericalMediumNCERT Intext Question 1.10
At a hill station, pure water boils at $96.80^{\circ} C$ . How much glucose ( $180 g m o l^{- 1}$ ) must be dissolved in $250 g$ of water to make it boil at $97.00^{\circ} C$ there? Take $K_{b} = 0.52 K k g m o l^{- 1}$ .
g
Show solution
$Δ T_{b} = 0.20 K$ , so $m = 0.20 / 0.52 = 0.385 m o l k g^{- 1}$ . In $0.250 k g$ of water: $n = 0.0962 m o l$ , mass $= 0.0962 \times 180 = 17.3 g$ of glucose.
NumericalMediumNCERT Intext Question 1.11
What mass of naphthalene ( $128 g m o l^{- 1}$ ) must be dissolved in $60.0 g$ of benzene to lower its freezing point by $0.80 K$ ? $K_{f} (benzene) = 5.12 K k g m o l^{- 1}$ .
g
Show solution
$m = 0.80 / 5.12 = 0.156 m o l k g^{- 1}$ ; $n = 0.156 \times 0.0600 = 9.38 \times 10^{- 3} m o l$ ; mass $= 9.38 \times 10^{- 3} \times 128 = 1.20 g$ .
NumericalMediumNCERT Intext Question 1.12
Calculate the osmotic pressure, in pascals, of a solution of $2.0 g$ of a polymer of molar mass $1.6 \times 10^{5} g m o l^{- 1}$ in $500 m L$ of water at $27^{\circ} C$ . Use $R = 8.314 J m o l^{- 1} K^{- 1}$ .
Pa
Show solution
$n = 2.0 / (1.6 \times 10^{5}) = 1.25 \times 10^{- 5} m o l$ in $5.0 \times 10^{- 4} m^{3}$ , so $C = 0.025 m o l m^{- 3}$ .
$Π = C R T = 0.025 \times 8.314 \times 300 = 62 P a$ : a tiny pressure, but still measurable as a height of about 6 mm of water.

1.7 Abnormal Molar Masses

Colligative properties count particles, and that can mislead. A student dissolves $1.00 g$ of sodium chloride in $99.0 g$ of water, measures a freezing-point depression of about $0.62 K$ , puts it into equation (1.36) and gets a molar mass near $30 g m o l^{- 1}$ , about half the formula mass of 58.5. Nothing was wrong with the measurement. In water each NaCl unit splits into a $N a^{+}$ and a $C l^{-}$ ion, so the solution holds nearly twice as many particles as formula units, the depression is nearly doubled, and the molar mass worked out from it is nearly halved. In general, dissociation into ions makes the measured molar mass smaller than the true one.

Molecules can also team up, which has the opposite effect. Ethanoic acid dissolved in benzene, a non-polar solvent of low dielectric constant that cannot hydrogen-bond to the acid, pairs up into dimers, each held together by two hydrogen bonds: $2 C H_{3} C O O H$ ⇌ $(C H_{3} C O O H)_{2}$ .

Complete dimerisation would halve the number of particles. Then $Δ T_{b}$ and $Δ T_{f}$ would be half the values expected for single molecules, and the molar mass worked out from them would come to about $120$ instead of $60 g m o l^{- 1}$ . A molar mass obtained from a colligative property that is lower or higher than the formula mass is called an abnormal molar mass. Fig. 1.14 compares the three cases.

2/3Each unit splits into two ions: 4 units give 8 particles, $i \approx 2$ .

Fig. 1.14Colligative properties count particles, so splitting raises the count and pairing lowers it.NCERT p. 24

Four formula units on the left are dissolved, and the particles they give are shown on the right, with the van't Hoff factor. In step 1, four glucose molecules in water stay as four particles: i equals 1. In step 2, four units of sodium chloride in water split into eight ions, four plus and four minus: i is about 2. In step 3, four ethanoic acid molecules in benzene pair up into two dimers, each pair joined by two dashed hydrogen bonds: i is about 0.5.

In the 1880s Jacobus van't Hoff introduced a correction factor $i$ , now called the van't Hoff factor, to correct for dissociation or association. It can be defined in three equivalent ways:

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}

Here the abnormal molar mass is the value found by experiment, and the 'calculated' colligative property is the one expected if the solute neither dissociates nor associates. For dissociation $i > 1$ ; for association $i < 1$ . Aqueous NaCl or KCl gives $i$ close to 2 when dilute, while ethanoic acid in benzene gives $i$ close to 0.5. Including $i$ , the colligative equations become:

Definition

van't Hoff factor

The ratio of the observed colligative property to the value calculated for an unchanged solute; equivalently, moles of particles present per mole of solute dissolved.

Colligative properties with the van't Hoff factor

\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 1.7Van't Hoff factors expected for complete dissociation or associationNCERT Table 1.4 · p. 25

Solute (solvent)	What happens	Ideal $i$
Glucose, urea (water)	molecules stay whole	1
$N a C l$ , $M g S O_{4}$ (water)	two ions per formula unit	2
$C a C l_{2}$ , $K_{2} S O_{4}$ (water)	three ions per formula unit	3
$A l C l_{3}$ (water)	four ions per formula unit	4
Ethanoic acid (benzene)	pairs of molecules (dimers)	0.5

Measured values for strong electrolytes fall a little short of these ideal numbers except in very dilute solutions. Oppositely charged ions attract one another, and some move around as close pairs that count as one particle. The shortfall is larger for ions with higher charges, such as $M g^{2 +}$ and $S O_{4}^{2 -}$ , and it shrinks as the solution is diluted.

DerivationDegree of dissociation or association from i

Dissociation. Let one mole of a solute that would give $n$ particles per formula unit be dissolved, and let a fraction $α$ (the degree of dissociation) actually dissociate. Then $1 - α$ mol stay whole and $n α$ mol of ions form:

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

Association. If $n$ molecules join to form one larger particle and a fraction $α$ (the degree of association) does so, $1 - α$ mol stay single and $α / n$ mol of associated particles form:

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

For dimerisation ( $n = 2$ ) this becomes $i = 1 - α / 2$ , so $α = 2 (1 - i)$ .

Example 1.12NCERT Example 1.12 · p. 25

How much ethanoic acid dimerises in benzene?

$1.20 g$ of ethanoic acid ( $C H_{3} C O O H$ , $60 g m o l^{- 1}$ ) is dissolved in $50.0 g$ of benzene, and the freezing point falls by $1.08 K$ . $K_{f} (benzene) = 5.12 K k g m o l^{- 1}$ . Find the observed molar mass, the van't Hoff factor and the percentage of the acid that is dimerised.

Try it yourself first. Then reveal the solution one step at a time.

Example 1.13NCERT Example 1.13 · p. 26

Degree of dissociation and Ka of a weak acid

$0.460 g$ of methanoic (formic) acid, $H C O O H$ ( $46 g m o l^{- 1}$ ), is dissolved in $500 g$ of water. The freezing point is lowered by $0.0406 K$ . Taking $K_{f} = 1.86 K k g m o l^{- 1}$ , find the van't Hoff factor, the degree of dissociation and the acid dissociation constant $K_{a}$ . In such a dilute aqueous solution, molality and molarity are practically equal.

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Quick check

1 of 4

A solute's molar mass measured by freezing-point depression in water comes out at about one third of its formula mass. The solute is most likely…

C a C l_{2}

N a C l

glucoseethanoic acid in benzene

Show answer

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

A molar mass one third of normal means

i \approx 3

: three particles per formula unit, as for

C a^{2 +}

+ 2

C l^{-}

. NaCl gives about 2, glucose 1, and an associating acid less than 1.

Recap

Solutions: key points and summary

A solution is a homogeneous mixture; the solvent sets its state, giving gaseous, liquid and solid solutions (Section 1.1).
Concentration can be stated as mass %, volume %, mass by volume %, ppm, mole fraction, molarity or molality; only the volume-based ones change with temperature (Section 1.2).
A saturated solution is a dynamic equilibrium; Le Chatelier's principle predicts how temperature changes the solubility of a solid (Section 1.3.1).
Henry's law, $p = K_{H} x$ : gas solubility rises with pressure and falls with temperature; a larger $K_{H}$ means a less soluble gas (Section 1.3.2).
Raoult's law, $p_{i} = p_{i}^{\circ} x_{i}$ ; for two volatile liquids the total vapour pressure is linear in composition and the vapour is richer in the more volatile component (Section 1.4.1).
Raoult's law is the special case of Henry's law with $K_{H} = p^{\circ}$ (Section 1.4.2).
A non-volatile solute lowers the vapour pressure of the solvent to $x_{1} p_{1}^{\circ}$ (Section 1.4.3).
Ideal solutions obey Raoult's law with $Δ_{mix} H = 0$ and $Δ_{mix} V = 0$ ; weaker A–B attractions give positive deviations, stronger ones negative (Section 1.5).
Large deviations give minimum- or maximum-boiling azeotropes, which distillation cannot separate (Section 1.5.2).
Colligative properties depend on the number of solute particles: 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).
Each colligative property gives the molar mass of a solute; osmotic pressure suits proteins and polymers (Section 1.6.4).
A pressure above the osmotic pressure drives reverse osmosis, used to purify water (Section 1.6.5).
The van't Hoff factor $i$ corrects for dissociation ( $i > 1$ ) and association ( $i < 1$ ) and gives the degree of either (Section 1.7).

Practice · 46 questions

Practice questions with answers (NCERT “Exercises”)

1
Short answerEasyNCERT Exercises 1.1
Answer both parts.
1. What makes a mixture a solution rather than a suspension? Which component is called the solvent?
  Show solution
  AnswerA solution is homogeneous down to the molecular scale; the solvent is the component (usually the most abundant) that sets its physical state.
  A solution has the same composition and properties throughout, because its components are mixed as individual molecules or ions; a suspension has particles big enough to settle out. The solvent is the component that decides the state of the solution, usually the one present in the largest amount; all others are solutes.
2. (b)
  Match each solution with its type (solute in solvent).
  Tap an item on the left, then its match on the right.
  Show answer
  Answer
  tincture of iodine : solid in liquid
  LPG vapour (propane with butane) : gas in gas
  sterling silver (copper in silver) : solid in solid
  gold amalgam (mercury in gold) : liquid in solid
  oxygen in a fish tank : gas in liquid
  Iodine dissolves in alcohol; propane and butane are both gases; copper in silver is a solid alloy; mercury, a liquid, dissolves in solid gold; oxygen gas dissolves in water.
Quick answerEasyNCERT Exercises 1.2
Pick the example in which a gas is dissolved in a solid.
Hydrogen absorbed into the metal of a hydride storage canisterAir bubbles trapped in a block of iceCarbon dioxide in soda waterSmoke particles in air
Show answer
Answer (A) Hydrogen absorbed into the metal of a hydride storage canister
Hydrogen molecules split into atoms that spread evenly through the metal lattice: a homogeneous solid with a gaseous solute. Bubbles in ice are a separate phase, soda water is a liquid solution, and smoke is a suspension of solid specks in a gas.
3
Short answerEasyNCERT Exercises 1.3
Concentration terms.
1. (a)
  Match each term with its definition.
  Tap an item on the left, then its match on the right.
  Show answer
  Answer
  mole fraction : moles of a component ÷ total moles
  molality : moles of solute ÷ kg of solvent
  molarity : moles of solute ÷ litres of solution
  mass percentage : mass of a component ÷ mass of solution × 100
  These are equations (1.4), (1.9), (1.8) and (1.1).
2. Which of these four measures changes when the solution is heated, and why?
  Show solution
  AnswerOnly molarity, because the volume of the solution grows on heating while masses and moles do not.
  Molarity has the volume of solution in its denominator; liquids expand when warmed, so the molarity falls slightly. Mole fraction, molality and mass percentage contain only masses or moles, which do not change with temperature.
NumericalMediumNCERT Exercises 1.4
Laboratory concentrated sulfuric acid is 98.0 % $H_{2} S O_{4}$ by mass and has a density of $1.84 g m L^{- 1}$ . Calculate its molarity ( $M (H_{2} S O_{4}) = 98.1 g m o l^{- 1}$ ).
M
Show solution
One litre of acid has a mass of $1000 \times 1.84 = 1840 g$ , of which $0.980 \times 1840 = 1803 g$ is $H_{2} S O_{4}$ .
Moles $= 1803 / 98.1 = 18.4 m o l$ in $1 L$ , so the acid is $18.4 M$ .
5
NumericalMediumNCERT Exercises 1.5
An aqueous urea solution ( $C O (N H_{2})_{2}$ , $60.06 g m o l^{- 1}$ ) is labelled 15.0 % w/w. Its density is $1.04 g m L^{- 1}$ .
1. Find its molality.
  mol/kg
  Show solution
  $n = 15.0 / 60.06 = 0.2498 m o l$ in $0.0850 k g$ of water: $m = 2.94 m o l k g^{- 1}$ .
2. Find the mole fraction of urea (the water's is 1 minus this).
  Show solution
  Water: $85.0 / 18.02 = 4.717 m o l$ . $x_{urea} = 0.2498 / (0.2498 + 4.717) = 0.0503$ ; $x_{water} = 0.950$ .
3. Find its molarity.
  M
  Show solution
  $V = 100 / 1.04 = 96.2 m L$ ; molarity $= 0.2498 / 0.0962 = 2.60 M$ .
NumericalHardNCERT Exercises 1.6
A $1.50 g$ sample is a mixture of potassium carbonate ( $K_{2} C O_{3}$ , $138.2 g m o l^{- 1}$ ) and potassium hydrogencarbonate ( $K H C O_{3}$ , $100.1 g m o l^{- 1}$ ) containing equal numbers of moles of each. What volume of $0.250 M$ HCl is needed to react with it completely?
mL
Show solution
Let each be $n$ mol: $138.2 n + 100.1 n = 1.50$ , so $n = 6.29 \times 10^{- 3} m o l$ .
Each carbonate ion takes two $H^{+}$ ions (forming water and carbon dioxide), and each hydrogencarbonate ion takes one.
HCl needed $= 2 n + n = 3 n = 0.0189 m o l$ . Volume $= 0.0189 / 0.250 = 0.0755 L = 75.5 m L$ .
NumericalEasyNCERT Exercises 1.7
$250 g$ of a 12.0 % (by mass) sugar syrup is mixed with $150 g$ of a 30.0 % syrup. What is the mass percentage of sugar in the mixture?
%
Show solution
Sugar: $0.120 \times 250 + 0.300 \times 150 = 30.0 + 45.0 = 75.0 g$ in $400 g$ of mixture, so $75.0 / 400 \times 100 = 18.75 %$ (about 18.8 %).
8
NumericalMediumNCERT Exercises 1.8
An engine coolant is made from $152 g$ of propylene glycol ( $C_{3} H_{8} O_{2}$ , $76 g m o l^{- 1}$ ) and $250 g$ of water. Its density is $1.035 g m L^{- 1}$ .
1. Calculate the molality.
  mol/kg
  Show solution
  $n = 152 / 76 = 2.00 m o l$ ; $m = 2.00 / 0.250 = 8.00 m o l k g^{- 1}$ .
2. Calculate the molarity.
  M
  Show solution
  Mass of solution $= 402 g$ ; $V = 402 / 1.035 = 388 m L$ ; molarity $= 2.00 / 0.388 = 5.15 M$ .
9
NumericalMediumNCERT Exercises 1.9
A groundwater sample contains 25 ppm (by mass) of nitrate ions, $N O_{3}^{-}$ ( $62 g m o l^{- 1}$ ).
1. Express this as a mass percentage.
  %
  Show solution
  $25 / 10^{6} \times 100 = 0.0025 %$ .
2. Find the molality of nitrate.
  mol/kg
  Show solution
  $25 g$ is $25 / 62 = 0.403 m o l$ in practically $10^{6} g = 1000 k g$ of water, so $m = 4.0 \times 10^{- 4} m o l k g^{- 1}$ .
Short answerMediumNCERT Exercises 1.10
Hexane is stirred into ethanol. Describe the intermolecular attractions before and after mixing, and predict the type of deviation from Raoult's law and the sign of $Δ_{mix} H$ .
Show solution
AnswerPositive deviation, with $Δ_{mix} H > 0$ .
Pure ethanol molecules are linked by hydrogen bonds; pure hexane has only weak London (dispersion) forces. On mixing, hexane molecules get between ethanol molecules and break many hydrogen bonds, offering only weak ethanol–hexane attractions in return.
Each kind of molecule is held less firmly than in its pure liquid, so it escapes more easily: the vapour pressure is above the Raoult's-law value (positive deviation). Energy is needed to break the hydrogen bonds, so mixing absorbs heat ( $Δ_{mix} H > 0$ ) and the mixture cools.
Short answerMediumNCERT Exercises 1.11
On a hot, still afternoon, fish in a shallow pond crowd near the surface. Use the solubility of gases to explain why.
Show solution
AnswerWarm water holds less dissolved oxygen, because dissolving a gas is exothermic.
Dissolving a gas is like condensing it, so heat is released ( $Δ_{sol} H < 0$ ). By Le Chatelier's principle, raising the temperature shifts the equilibrium $gas + water ⇌ solution$ back towards free gas, so less oxygen stays dissolved ( $K_{H}$ grows with temperature).
The warm pond therefore holds less oxygen, and the fish move to the surface layer, where oxygen from the air is replenished fastest.
12
Short answerEasyNCERT Exercises 1.12
State Henry's law and use it to explain each observation.
1. State Henry's law.
  Show solution
  AnswerAt constant temperature, the partial pressure of a gas over a solution is proportional to its mole fraction in the solution: $p = K_{H} x$ .
  At a fixed temperature, $p = K_{H} x$ : the more gas pressure above the liquid, the more gas dissolves.
2. A soda bottle fizzes when opened.
  Show solution
  It was sealed under a high CO2 pressure, so a lot of gas dissolved. Opening drops the pressure to that of air, so the equilibrium mole fraction falls and the excess gas escapes as bubbles.
3. Divers must come up slowly.
  Show solution
  At depth they breathe air at high pressure, so extra nitrogen dissolves in the blood. Rising quickly lowers the pressure suddenly and the nitrogen forms bubbles in the blood (the bends); rising slowly lets it leave through the lungs.
4. Climbers carry oxygen on very high peaks.
  Show solution
  The partial pressure of oxygen is low at altitude, so little oxygen dissolves in the blood (anoxia); breathing oxygen-enriched gas raises its partial pressure.
NumericalEasyNCERT Exercises 1.13
At a fixed temperature, $2.4 \times 10^{- 2} g$ of methane dissolves in a sample of a solvent when the methane pressure is $0.80 b a r$ . What partial pressure is needed to dissolve $6.0 \times 10^{- 2} g$ in the same amount of solvent?
bar
Show solution
By Henry's law $p \propto x$ , and for small amounts $x$ is proportional to the mass dissolved. So $p = 0.80 \times (6.0 / 2.4) = 2.0 b a r$ .
14
Short answerMediumNCERT Exercises 1.14
Deviations and heat of mixing.
1. Mixing liquids X and Y absorbs heat and the volume increases slightly. What deviation from Raoult's law do you expect?
  Positive deviationNegative deviationNo deviation
  Show answer
  Answer (A) Positive deviation
  Heat absorbed means weaker X–Y attractions than X–X and Y–Y: molecules escape more easily and the vapour pressure is higher than ideal.
2. Mixing liquids U and V releases heat. Which statement fits?
  U–V attractions are weaker; vapour pressure above RaoultU–V attractions are stronger; vapour pressure below RaoultThe mixture must be ideal
  Show answer
  Answer (B) U–V attractions are stronger; vapour pressure below Raoult
  Heat released ( $Δ_{mix} H < 0$ ) means stronger U–V attractions, a lower escaping tendency and a negative deviation.
3. Explain in one or two sentences what positive and negative deviations mean.
  Show solution
  A positive deviation means the measured vapour pressure is higher than Raoult's law predicts (A–B weaker, $Δ_{mix} H > 0$ ); a negative deviation means it is lower (A–B stronger, $Δ_{mix} H < 0$ ).
NumericalHardNCERT Exercises 1.15
A 3.0 % (by mass) aqueous solution of a non-volatile non-electrolyte has a vapour pressure of $1.003 b a r$ at $373.15 K$ , where pure water's is $1.013 b a r$ . Find the molar mass of the solute.
g/mol
Show solution
Relative lowering: $x_{2} = (1.013 - 1.003) / 1.013 = 0.00987$ .
In $100 g$ : $97.0 g$ of water $= 5.38 m o l$ . From $x_{2} = n_{2} / (n_{1} + n_{2})$ : $n_{2} = x_{2} n_{1} / (1 - x_{2}) = 0.0537 m o l$ .
$M_{2} = 3.0 / 0.0537 = 56 g m o l^{- 1}$ . (The dilute approximation $n_{2} \approx x_{2} n_{1}$ gives $56.5$ , the same to two significant figures.)
16
NumericalMediumNCERT Exercises 1.16
Hexane ( $86 g m o l^{- 1}$ ) and heptane ( $100 g m o l^{- 1}$ ) form an ideal solution. At $333 K$ their vapour pressures are $76.0 k P a$ and $28.0 k P a$ . A mixture contains $43.0 g$ of hexane and $50.0 g$ of heptane.
1. What is its vapour pressure at $333 K$ ?
  kPa
  Show solution
  $n = 0.500 m o l$ of each, so $x = 0.500$ . $p = 0.500 \times 76.0 + 0.500 \times 28.0 = 52.0 k P a$ .
2. What is the mole fraction of hexane in the vapour?
  Show solution
  $y_{hexane} = 38.0 / 52.0 = 0.731$ .
NumericalMediumNCERT Exercises 1.17
Benzene ( $78 g m o l^{- 1}$ ) has a vapour pressure of $12.7 k P a$ at $298 K$ . Calculate the vapour pressure of a $0.500 m o l k g^{- 1}$ solution of a non-volatile solute in benzene at this temperature.
kPa
Show solution
$1 k g$ of benzene $= 1000 / 78 = 12.82 m o l$ , with $0.500 m o l$ of solute: $x_{2} = 0.500 / 13.32 = 0.0375$ .
$p = 12.7 \times (1 - 0.0375) = 12.2 k P a$ .
NumericalMediumNCERT Exercises 1.18
What mass of a non-volatile solute of molar mass $60 g m o l^{- 1}$ must be dissolved in $92.0 g$ of toluene ( $92 g m o l^{- 1}$ ) to lower its vapour pressure to 90.0 % of the pure value?
g
Show solution
$p = x_{1} p^{\circ}$ , so $x_{1} = 0.900$ . Toluene: $n_{1} = 1.00 m o l$ .
$n_{2} = n_{1} (1 - x_{1}) / x_{1} = 1.00 \times 0.100 / 0.900 = 0.111 m o l$ , so mass $= 0.111 \times 60 = 6.67 g$ .
19
NumericalHardNCERT Exercises 1.19
A solution of $20.0 g$ of a non-volatile solute in $72.0 g$ of water shows a vapour pressure of $3.01 k P a$ at a certain temperature. After $36.0 g$ more water is added, the vapour pressure is $3.07 k P a$ at the same temperature. Find (a) the molar mass of the solute and (b) the vapour pressure of pure water at this temperature.
1. Molar mass of the solute
  g/mol
  Show solution
  Water: $72.0 g = 4.00 m o l$ , then $108 g = 6.00 m o l$ . Call the solute $n$ mol.
  First solution: $3.01 = 4.00 p^{\circ} / (4.00 + n)$ .
  Second solution: $3.07 = 6.00 p^{\circ} / (6.00 + n)$ .
  Dividing the first by the second removes $p^{\circ}$ and gives $n = 0.249 m o l$ , so $M = 20.0 / 0.249 = 80 g m o l^{- 1}$ .
2. Vapour pressure of pure water
  kPa
  Show solution
  $p^{\circ} = 3.01 \times (4.00 + 0.249) / 4.00 = 3.20 k P a$ .
NumericalMediumNCERT Exercises 1.20
A 4.0 % (by mass) aqueous urea solution ( $60 g m o l^{- 1}$ ) freezes at $271.86 K$ . Without using $K_{f}$ , find the freezing point of a 4.0 % aqueous glucose solution ( $180 g m o l^{- 1}$ ). Pure water freezes at $273.15 K$ .
K
Show solution
Both solutions have $4.0 g$ of solute in $96.0 g$ of water, so their molalities are in the inverse ratio of the molar masses, $60 : 180 = 1 : 3$ .
$Δ T_{f} (urea) = 273.15 - 271.86 = 1.29 K$ , so $Δ T_{f} (glucose) = 1.29 / 3 = 0.43 K$ and the glucose solution freezes at $272.72 K$ .
NumericalHardNCERT Exercises 1.21
Two compounds of elements X and Y have formulae $X Y_{3}$ and $X Y_{5}$ . Dissolved in $25.0 g$ of benzene ( $K_{f} = 5.12 K k g m o l^{- 1}$ ), $1.00 g$ of $X Y_{3}$ lowers the freezing point by $1.49 K$ and $1.00 g$ of $X Y_{5}$ lowers it by $0.984 K$ . Find the atomic masses of X and Y, and suggest what the elements are.
Show solution
Answer $M (X Y_{3}) = 137$ , $M (X Y_{5}) = 208$ ; Y ≈ 35.3 (chlorine), X ≈ 31.4 (phosphorus).
$M (X Y_{3}) = 5.12 \times 1.00 \times 1000 / (1.49 \times 25.0) = 137.4 g m o l^{- 1}$ ; $M (X Y_{5}) = 5.12 \times 1000 / (0.984 \times 25.0) = 208.1 g m o l^{- 1}$ .
$2 Y = 208.1 - 137.4 = 70.7$ , so $Y = 35.3$ . Then $X = 137.4 - 3 \times 35.3 = 31.4$ .
These fit chlorine (35.5) and phosphorus (31.0): the compounds are $P C l_{3}$ and $P C l_{5}$ .
NumericalMediumNCERT Exercises 1.22
At $310 K$ , a solution containing $18.0 g$ of glucose per litre has an osmotic pressure of $2.58 b a r$ . What glucose concentration gives $7.74 b a r$ , close to the osmotic pressure of blood, at the same temperature?
M
Show solution
$18.0 g L^{- 1}$ of glucose is $0.100 M$ . Since $Π = C R T$ , $C_{2} = C_{1} Π_{2} / Π_{1} = 0.100 \times 7.74 / 2.58 = 0.300 M$ ( $54.0 g L^{- 1}$ ).
23
Quick answerMediumNCERT Exercises 1.23
Match each pair of substances with the most important attraction between them.
Tap an item on the left, then its match on the right.
Show answer
Answer
pentane and hexane : London (dispersion) forces
potassium bromide and water : ion–dipole attraction
ethanol and water : hydrogen bonding
acetone and ethyl ethanoate : dipole–dipole attraction
Alkanes are non-polar; ions are attracted to the polar water molecules; both ethanol and water have O–H groups; acetone and ethyl ethanoate are polar but have no O–H or N–H, so they cannot hydrogen-bond with each other.
24
Quick answerMediumNCERT Exercises 1.24
Arrange these in order of increasing solubility in hexane: pentane, sodium bromide, acetone, ethylene glycol.
1. acetone
2. ethylene glycol
3. pentane
4. sodium bromide
Show answer
Answer
sodium bromide
ethylene glycol
acetone
pentane
An ionic solid barely dissolves in a non-polar liquid. Ethylene glycol, with two O–H groups, is strongly hydrogen-bonded and only slightly soluble. Acetone is polar but mixes with hexane. Pentane, a non-polar alkane like hexane, mixes in all proportions.
25
Quick answerMediumNCERT Exercises 1.25
Match each compound with its solubility in water at room temperature.
Tap an item on the left, then its match on the right.
Show answer
Answer
hexane : insoluble
1-butanol : partially soluble
glycerol : highly soluble (miscible)
Hexane has no polar group. 1-Butanol's O–H dissolves some of it (about 8 g per 100 g of water), but the four-carbon chain limits this. Glycerol has three O–H groups on a small molecule and mixes completely.
NumericalHardNCERT Exercises 1.26
A sample of brine contains $69.0 g$ of $N a^{+}$ ions (with the matching chloride, as NaCl) for every $1.00 k g$ of water. Its density is $1.11 g m L^{- 1}$ . Calculate the molarity of $N a^{+}$ .
M
Show solution
$n (N a^{+}) = 69.0 / 22.99 = 3.00 m o l$ , with $3.00 \times 35.45 = 106.4 g$ of $C l^{-}$ .
Mass of brine $= 1000 + 69.0 + 106.4 = 1175 g$ ; volume $= 1175 / 1.11 = 1059 m L$ .
Molarity of $N a^{+}$ $= 3.00 / 1.059 = 2.83 M$ .
NumericalMediumNCERT Exercises 1.27
The solubility product of silver bromide, $A g B r$ , is $5.0 \times 10^{- 13}$ at $298 K$ . What is the largest molarity of AgBr that can exist in water at this temperature?
M
Show solution
Each AgBr unit that dissolves gives one $A g^{+}$ and one $B r^{-}$ ion. With solubility $s$ , $[A g^{+}] = [B r^{-}] = s$ and $K_{s p} = s^{2}$ .
$s = \sqrt{5.0 \times 10^{- 13}} = 7.1 \times 10^{- 7} m o l L^{- 1}$ (about $0.13 m g$ per litre).
NumericalEasyNCERT Exercises 1.28
$2.5 g$ of paracetamol ( $C_{8} H_{9} N O_{2}$ ) is dissolved in $75 g$ of ethanol. Calculate the mass percentage of paracetamol.
%
Show solution
$2.5 / (2.5 + 75) \times 100 = 3.23 %$ .
NumericalMediumNCERT Exercises 1.29
A patient is to receive $50 m g$ of caffeine ( $C_{8} H_{10} N_{4} O_{2}$ , $194.2 g m o l^{- 1}$ ) from a $2.0 \times 10^{- 3} m o l k g^{- 1}$ aqueous solution. What mass of the solution is needed?
g
Show solution
$n = 0.050 / 194.2 = 2.57 \times 10^{- 4} m o l$ . Water needed $= n / m = 2.57 \times 10^{- 4} / 2.0 \times 10^{- 3} = 0.129 k g$ .
Adding the $0.05 g$ of caffeine changes nothing at this precision: about $129 g$ of solution.
NumericalEasyNCERT Exercises 1.30
What mass of salicylic acid ( $C_{7} H_{6} O_{3}$ , $138.1 g m o l^{- 1}$ ) is needed to make $500 m L$ of a $0.120 M$ solution in ethanol?
g
Show solution
$n = 0.120 \times 0.500 = 0.0600 m o l$ ; mass $= 0.0600 \times 138.1 = 8.29 g$ .
31
Quick answerMediumNCERT Exercises 1.31
Equal molalities of ethanoic acid, chloroethanoic acid and dichloroethanoic acid are dissolved in water. Arrange them from the smallest to the largest freezing-point depression.
1. chloroethanoic acid
2. ethanoic acid
3. dichloroethanoic acid
Show answer
Answer
ethanoic acid
chloroethanoic acid
dichloroethanoic acid
Each chlorine pulls electron density away from the –COOH group and stabilises the anion, so the acid strength rises with the number of chlorines. A stronger acid dissociates more, giving a larger $i$ and a larger $Δ T_{f}$ .
NumericalHardNCERT Exercises 1.32
$8.00 g$ of chloroethanoic acid ( $C l C H_{2} C O O H$ , $94.5 g m o l^{- 1}$ , $K_{a} = 1.35 \times 10^{- 3}$ ) is dissolved in $400 g$ of water. Estimate the depression of the freezing point ( $K_{f} = 1.86 K k g m o l^{- 1}$ ).
K
Show solution
$m = (8.00 / 94.5) / 0.400 = 0.212 m o l k g^{- 1}$ ; take $c \approx 0.212 M$ .
Solve $0.212 α^{2} + 1.35 \times 10^{- 3} α - 1.35 \times 10^{- 3} = 0$ : $α = 0.077$ , so $i = 1 + α = 1.077$ .
$Δ T_{f} = i K_{f} m = 1.077 \times 1.86 \times 0.212 = 0.424 K$ (compared with $0.394 K$ if the acid did not dissociate).
33
NumericalHardNCERT Exercises 1.33
$13.9 g$ of bromoethanoic acid ( $B r C H_{2} C O O H$ , $138.9 g m o l^{- 1}$ ) is dissolved in $500 g$ of water, and the freezing point falls by $0.401 K$ . Find $i$ for the acid, and from it the acid's $K_{a}$ ( $K_{f} = 1.86 K k g m o l^{- 1}$ ).
1. van't Hoff factor
  Show solution
  $13.9 g$ is $0.100 m o l$ , so $m = 0.200 m o l k g^{- 1}$ .
  Calculated $Δ T = 1.86 \times 0.200 = 0.372 K$ , so $i = 0.401 / 0.372 = 1.078$ .
2. Dissociation constant
  Show solution
  $α = i - 1 = 0.078$ ; $K_{a} = 0.200 \times {0.078}^{2} / (1 - 0.078) = 1.3 \times 10^{- 3}$ .
NumericalMediumNCERT Exercises 1.34
The vapour pressure of water at $308 K$ is $42.2 m m H g$ . Calculate the vapour pressure of a solution of $36.0 g$ of glucose ( $180 g m o l^{- 1}$ ) in $500 g$ of water at this temperature.
mmHg
Show solution
$n_{2} = 0.200 m o l$ , $n_{1} = 500 / 18.02 = 27.75 m o l$ ; $x_{1} = 27.75 / 27.95 = 0.9928$ .
$p = 0.9928 \times 42.2 = 41.9 m m H g$ .
35
NumericalMediumNCERT Exercises 1.35
Henry's law constant for nitrogen in water at $298 K$ is $6.5 \times 10^{7} m m H g$ . Find the mole fraction and the molality of nitrogen in water exposed to nitrogen at a partial pressure of $593 m m H g$ (its share of the air at sea level).
1. Mole fraction
  Show solution
  $x = 593 / 6.5 \times 10^{7} = 9.1 \times 10^{- 6}$ .
2. Molality
  mol/kg
  Show solution
  $n \approx x \times 55.5 = 5.1 \times 10^{- 4} m o l$ per kilogram of water.
36
NumericalMediumNCERT Exercises 1.36
$50.0 g$ of liquid P ( $100 g m o l^{- 1}$ ) is mixed with $400 g$ of liquid Q ( $200 g m o l^{- 1}$ ) to form an ideal solution. Pure Q has a vapour pressure of $300 t o r r$ , and the solution's total vapour pressure is $280 t o r r$ . Find the vapour pressure of pure P and its partial pressure in the solution.
1. Partial pressure of P in the solution
  torr
  Show solution
  $n_{P} = 0.500$ , $n_{Q} = 2.00$ , so $x_{P} = 0.200$ , $x_{Q} = 0.800$ . $p_{Q} = 0.800 \times 300 = 240 t o r r$ , so $p_{P} = 280 - 240 = 40 t o r r$ .
2. Vapour pressure of pure P
  torr
  Show solution
  $p_{P}^{\circ} = p_{P} / x_{P} = 40 / 0.200 = 200 t o r r$ .
DiagramHardNCERT Exercises 1.37
Pure liquids P and Q have vapour pressures of $520$ and $380 m m H g$ at $320 K$ . The measured partial pressures over their mixtures are given below. Plot $p_{P}$ , $p_{Q}$ and $p_{total}$ against $x_{P}$ , add the ideal (Raoult's law) lines, and decide which kind of deviation the mixture shows.
Show solution
AnswerPositive deviation: every measured total lies above the ideal line; the total even passes 520 mmHg near $x_{P} = 0.8$ , so a minimum-boiling azeotrope forms near there.
Table 1.8Measured pressures (mmHg) at 320 K, with the ideal total for comparison
$x_{P}$ $p_{P}$ $p_{Q}$ 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
The ideal values come from $p_{P} = 520 x_{P}$ and $p_{Q} = 380 (1 - x_{P})$ . Every measured partial pressure and total is above its ideal value, so the mixture shows a positive deviation: P–Q attractions are weaker than P–P and Q–Q.
Near $x_{P} = 0.8$ the measured total ( $538 m m H g$ ) exceeds even pure P's $520 m m H g$ , so the total curve has a maximum. A mixture of about this composition is a minimum-boiling azeotrope.
NumericalMediumNCERT Exercises 1.38
Toluene ( $92 g m o l^{- 1}$ ) and ethylbenzene ( $106 g m o l^{- 1}$ ) form an ideal solution. At $360 K$ their vapour pressures are $48.9 k P a$ and $21.6 k P a$ . Find the mole fraction of toluene in the vapour above a mixture of $46.0 g$ of toluene and $106 g$ of ethylbenzene.
Show solution
$n_{tol} = 0.500 m o l$ , $n_{eb} = 1.00 m o l$ , so $x_{tol} = 0.333$ .
$p_{tol} = 0.333 \times 48.9 = 16.3 k P a$ ; $p_{eb} = 0.667 \times 21.6 = 14.4 k P a$ ; total $30.7 k P a$ .
$y_{tol} = 16.3 / 30.7 = 0.531$ : a third of the liquid but more than half of the vapour.
39
NumericalHardNCERT Exercises 1.39
Air is about 21 % oxygen and 78 % nitrogen by volume. Water at $298 K$ is in equilibrium with air at a total pressure of $5.00 b a r$ . Using $K_{H} = 44 k b a r$ for $O_{2}$ and $86 k b a r$ for $N_{2}$ , find the mole fraction of each gas dissolved. Is the dissolved air richer or poorer in oxygen than the air above?
1. Mole fraction of oxygen
  Show solution
  $p (O_{2}) = 0.21 \times 5.00 = 1.05 b a r$ ; $x = 1.05 / 44 000 = 2.39 \times 10^{- 5}$ .
2. Mole fraction of nitrogen
  Show solution
  $p (N_{2}) = 0.78 \times 5.00 = 3.90 b a r$ ; $x = 3.90 / 86 000 = 4.53 \times 10^{- 5}$ .
3. Richer or poorer in oxygen?
  Show solution
  AnswerRicher.
  Dissolved N2 : O2 $= 4.53 / 2.39 = 1.9$ , against $78 / 21 = 3.7$ in the air. Oxygen is the more soluble gas (smaller $K_{H}$ ), so dissolved air is about twice as rich in oxygen as the atmosphere, which helps aquatic animals.
NumericalMediumNCERT Exercises 1.40
What mass of magnesium chloride ( $M g C l_{2}$ , $95.2 g m o l^{- 1}$ , $i = 2.60$ ) must be dissolved in water to make $1.50 L$ of solution with an osmotic pressure of $1.20 b a r$ at $300 K$ ? ( $R = 0.0831 L b a r m o l^{- 1} K^{- 1}$ )
g
Show solution
$n = Π V / (i R T) = 1.20 \times 1.50 / (2.60 \times 0.0831 \times 300) = 0.0278 m o l$ .
Mass $= 0.0278 \times 95.2 = 2.64 g$ .
NumericalMediumNCERT Exercises 1.41
Find the osmotic pressure at $37^{\circ} C$ of a solution of $40.0 m g$ of sodium sulfate ( $N a_{2} S O_{4}$ , $142.0 g m o l^{- 1}$ ) in $500 m L$ of water, assuming complete dissociation. ( $R = 0.0831 L b a r m o l^{- 1} K^{- 1}$ )
bar
Show solution
Each $N a_{2} S O_{4}$ gives two $N a^{+}$ ions and one $S O_{4}^{2 -}$ ion, so $i = 3$ .
$n = 0.0400 / 142.0 = 2.82 \times 10^{- 4} m o l$ in $0.500 L$ : $C = 5.63 \times 10^{- 4} M$ .
$Π = i C R T = 3 \times 5.63 \times 10^{- 4} \times 0.0831 \times 310 = 0.0436 b a r$ ( $4.36 k P a$ ).
Quick answerEasy
Which of these is not a colligative property?
Osmotic pressureElevation of boiling pointVapour pressure of the pure solventDepression of freezing point
Show answer
Answer (C) Vapour pressure of the pure solvent
The pure solvent's vapour pressure is a property of the solvent alone; the colligative property is the lowering of vapour pressure caused by a solute.
Quick answerMedium
A $0.100 m o l k g^{- 1}$ aqueous solution of a salt freezes at $- 0.502^{\circ} C$ . What is its van't Hoff factor? ( $K_{f} = 1.86 K k g m o l^{- 1}$ )
Show answer
Answer 2.7
Calculated $Δ T_{f} = 0.186 K$ ; $i = 0.502 / 0.186 = 2.7$ , suggesting a salt such as $C a C l_{2}$ that is not quite fully dissociated.
Quick answerEasy
Two solutions that have the same osmotic pressure at the same temperature are called solutions.
Show answer
Answer isotonic
Isotonic solutions show no net osmosis across a semipermeable membrane.
45
Quick answerMedium
Arrange these aqueous solutions, all $0.01 m o l k g^{- 1}$ , in order of increasing boiling point.
1. $K N O_{3}$
2. $N a_{2} S O_{4}$
3. glucose
4. $A l C l_{3}$
Show answer
Answer
glucose
$K N O_{3}$
$N a_{2} S O_{4}$
$A l C l_{3}$
Ideal $i$ values are 1, 2, 3 and 4, so the boiling-point elevation, and the boiling point, increase in this order.
Quick answerEasy
The molality of a solution changes when the solution is warmed.
TrueFalse
Show answer
Answer False
False. Molality is moles of solute per kilogram of solvent; neither changes with temperature. Molarity, which uses volume, does change.

$x_{P}$	$p_{P}$	$p_{Q}$	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

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NCERT Intext Question 1.1	Practice: concentration (Intext Questions 1.1–1.5) · Q1
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Quick check: MCQs on Solutions

1.1 Types of Solutions3 questions
Mole fraction, molarity and 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: A liquid mixture that boils at a constant temperature to give vapour of the same composition, so distillation cannot separate it.First seen in the chapter →

colligative propertyअणुसंख्य गुणधर्मanusankhya gundharm: A property of a dilute solution that depends on the number of solute particles, not on what they are.First seen in the chapter →

Henry's lawहेनरी का नियमHenry ka niyam: At constant temperature the partial pressure of a gas over a solution is proportional to its mole fraction in the solution, $p = K_{H} x$ .

ideal solutionआदर्श विलयनaadarsh vilayan: A solution that obeys Raoult's law at every composition, with no heat change and no volume change on mixing.First seen in the chapter →

molalityमोललताmolalta: Moles of solute per kilogram of solvent, in $m o l k g^{- 1}$ (m); it does not depend on temperature.First seen in the chapter →
molarityमोलरताmolarta: Moles of solute per litre of solution, in $m o l L^{- 1}$ (M); it changes slightly with temperature.First seen in the chapter →
mole fractionमोल अंशmol ansh: Moles of one component divided by the total moles of all components; the mole fractions of a solution add up to 1.First seen in the chapter →

osmotic pressureपरासरण दाबparasaran daab: The extra pressure that must be put on a solution to stop solvent flowing into it through a semipermeable membrane; $Π = C R T$ .First seen in the chapter →

Raoult's lawराउल्ट का नियमRaoult ka niyam: Every volatile component of a solution has a partial vapour pressure equal to its mole fraction times its pure vapour pressure, $p_{i} = x_{i} p_{i}^{\circ}$ .First seen in the chapter →

saturated solutionसंतृप्त विलयनsantript vilayan: A solution in dynamic equilibrium with undissolved solute, so it cannot dissolve any more at that temperature and pressure.First seen in the chapter →
solubilityविलेयताvileyata: The largest amount of a substance that dissolves in a stated amount of solvent at a given temperature and pressure.First seen in the chapter →
soluteविलेयvilay: Any component of a solution other than the solvent.First seen in the chapter →
solutionविलयनvilayan: A mixture that is uniform throughout, down to the scale of molecules, made of two or more components.First seen in the chapter →
solventविलायकvilaayak: The component of a solution that is usually present in the largest amount and decides its physical state.First seen in the chapter →

van't Hoff factorवान्ट हॉफ गुणकvant Hoff gunak: The ratio $i$ of the particles actually present to the formula units dissolved; above 1 for dissociation, below 1 for association.First seen in the chapter →

$x_{P}$	$p_{P}$	$p_{Q}$	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 and parts per million

Mole fraction, molarity and 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

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$x_{P}$	$p_{P}$	$p_{Q}$	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