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ChemistryCh 1 Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry: Class 11 Chemistry · Chapter 1

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

Published · Updated

  • 107 min
  • 1 simulation
  • 39 practice questions
  • 32 MCQs

What you will learn in Some Basic Concepts of Chemistry

  • Describe how chemistry developed, including India's long chemical tradition, and explain its importance in daily life and the economy
  • Explain the three states of matter in terms of particles and classify matter as mixtures, elements or compounds
  • Use SI base units and prefixes, distinguish mass from weight, relate density and temperature scales, and convert units by dimensional analysis
  • Write numbers in scientific notation, apply significant-figure and rounding rules, and distinguish precision from accuracy
  • State and apply the five laws of chemical combination and explain them with Dalton's atomic theory and Avogadro's law
  • Calculate atomic, average atomic, molecular and formula masses and use the mole, molar mass and Avogadro constant
  • Find mass percentages, empirical formulas and molecular formulas from composition data
  • Carry out stoichiometric calculations, including limiting reagents and concentrations as mass per cent, mole fraction, molarity and molality

Original companion text aligned to NCERT Chemistry Class 11, Chapter 1, pp. 1–28. Not affiliated with or endorsed by NCERT.

Leave dosa batter out overnight and by morning it has risen and turned sour. An iron gate left out through the monsoon grows a flaky brown crust. A green mango shut in a paper bag turns soft and sweet. In each case the substances themselves have changed into new ones. Chemistry (रसायन विज्ञान) asks what substances are made of, how their particles are arranged, how they behave and how they turn into other substances.

This unit builds the toolkit for the whole course. We sort matter into types, learn to measure quantities and report them honestly, meet the laws that persuaded scientists that atoms are real, and learn to count atoms by weighing them with the mole. By the end you will be able to predict how much product a reaction can give, a skill you will use in almost every later chapter.

Development of Chemistry

Chemistry as an organised, measuring science is only about three hundred years old, but people practised it long before that. Medieval alchemists chased two prizes that never existed: a philosopher's stone (पारस, paras) to turn iron or copper into gold, and an elixir of life. From about 1300 to 1600 CE many turned to medicines, a phase called iatrochemistry. The quests failed, but they left real skills behind: distilling, crystallising, purifying and weighing. Arab scholars carried this craft into Europe, where chemistry became a quantitative science in the eighteenth century.

China and India had alchemical traditions of their own. Sanskrit texts call the subject Rasayana Shastra; it covered metallurgy, medicine, dyes, glass and cosmetics. The timeline picks out a few Indian achievements for which the evidence is good.

Timeline

Chemistry in India through the ages

  1. c. 2600–1900 BCE

    Harappan crafts

    Kiln-fired bricks, glazed faience beads and copper hardened with tin or arsenic: all products of controlled heating.

  2. BCE (dates debated)

    Kanada's paramanu

    The Vaisheshika school of Kanada taught that substances are built from tiny, eternal particles, paramanu, which can join in pairs and triplets.

  3. From c. 300 BCE

    Wootz steel

    South Indian smiths made high-carbon wootz steel in sealed clay crucibles; it was traded far to the west.

  4. Early centuries CE

    Ayurvedic chemistry

    The Charaka and Sushruta Samhitas describe caustic alkalis (kshara) and bhasmas, metal powders made by repeated heating.

  5. c. 400 CE

    Delhi iron pillar

    A wrought-iron pillar over 7 m tall has stood in the open for about sixteen centuries with very little rust.

  6. 6th century CE

    Brihat Samhita

    Varahamihira's encyclopaedia gives recipes for perfumes and for a hard plaster (vajralepa) from boiled plant extracts and resins.

  7. Early medieval

    Rasaratnakara

    A treatise attributed to Nagarjuna describes mercury compounds and the extraction of copper, silver and gold.

  8. Medieval

    Zinc at Zawar

    Retorts at Zawar, Rajasthan, show zinc being distilled centuries before the method was used in Europe.

  9. 1892

    Bengal Chemicals

    Acharya Prafulla Chandra Ray began making medicines and chemicals in Kolkata, starting one of India's first modern chemical industries.

Some of these ideas sound surprisingly modern. Grinding and heating metals into very fine bhasma powders changes how they behave, and studies of some bhasmas report particles only tens of nanometres across, the size range of today's nanotechnology. Kanada's paramanu, proposed many centuries before John Dalton's atomic theory of 1808, is the clearest early Indian statement that matter is made of particles. From the eighteenth century, imported goods and colonial rule pushed many indigenous techniques into decline, and modern laboratory chemistry reached Indian colleges only in the late nineteenth century.

What these traditions lacked was a way to count the particles Kanada imagined. Before this unit ends you will be able to say how many molecules there are in a single paracetamol tablet.

Quick check
1 of 2
Which description fits iatrochemistry?
Show answer

Answer (B) Chemistry applied mainly to preparing medicines

Iatrochemists (roughly 1300–1600 CE) turned alchemy towards medicines. Gold-making was the older alchemical quest.

1.1 Importance of Chemistry

Chemistry is often called the central science: it borrows laws from physics and lends explanations to biology, geology, medicine and engineering. The ozone overhead, the signal racing along a nerve and the silicon chip in your phone are all chemistry at work.

  • Food. Fertilisers such as urea feed most of India's crops; since 2015 all Indian urea has been neem-coated so that its nitrogen is released more slowly. Pesticides, preservatives and packaging come from chemists too.
  • Health. Chemists design, make and purify medicines, from paracetamol to anti-cancer drugs. Oral rehydration solution, a balanced mix of glucose and salts, has saved millions of children with diarrhoea.
  • Materials. Polymers, optical fibres, lithium-ion batteries, superconducting ceramics and conducting plastics all came from designing substances with chosen electrical, magnetic or optical properties.
  • Environment. Chemists replaced the CFCs that were thinning the ozone layer, which is now slowly recovering. Managing greenhouse gases such as carbon dioxide and methane is a big open problem.
  • Economy. Industries making acids, alkalis, fertilisers, dyes, cement, polymers, metals and medicines employ millions of people.

Challenges remain, from the chemistry of living cells to storing solar energy and turning waste into resources. All of that work rests on the basic ideas of this unit, starting with what matter is.

Quick check
Which of these are achievements mainly of chemistry? Choose all that apply.
Select all that apply
Show answer
Answer
  • (A) Substitutes for ozone-damaging CFCs
  • (B) Neem-coated urea that releases nitrogen slowly
  • (D) Lithium-ion batteries
CFC substitutes, coated fertilisers and battery materials all depend on designing new substances. Measuring the Moon's distance is physics and astronomy.

1.2 Nature of Matter

A pebble, the air in a football, the water in a matka and your own body share two things: each has mass, and each takes up space. Anything with both properties is matter (द्रव्य). Light, heat and sound are not matter; they are forms of energy.

1.2.1 States of Matter

Whether a sample is a solid, a liquid or a gas depends on how its particles are spaced and how freely they move. Locked into a regular pattern and able only to jiggle on the spot: that is a solid. Still touching, but free to slide past one another: a liquid. Widely spaced and racing about at random: a gas.

Table 1.1How particle arrangement decides shape and volumeNCERT Fig. 1.1 · p. 4
ParticlesShapeVolume
Solidpacked, ordered, vibrate in placefixedfixed
Liquidclose, but slide past each othertakes the container's shapefixed
Gasfar apart, fast, randomfills the containerfills the container

Changing the temperature or the pressure turns one state into another. Heat ice and it melts; heat the water and it boils; cool the steam and the steps run backwards. Pressure matters too: LPG is a gas in the open air but is squeezed into a liquid inside a cylinder, which is why a full cylinder sloshes.

SimulationParticles in the three states of water

A closed box of water particles with a thermometer and a temperature slider from minus 20 to 120 degrees Celsius. Below 0 degrees the particles sit in a regular pattern and only vibrate; between 0 and 100 degrees they stay close but move past one another at the bottom; above 100 degrees they spread fast through the whole box. A label names the state.

Fig. 1.1Water particles in a closed box. Press Play, then move the temperature slider from ice to steam.
Try this
  • Start at −10 ∘C and press Play. Do the particles change places, or only shake?
  • Warm the box to 50 ∘C. Which changes: the sample's shape or its volume?
  • Go above 100 ∘C. Why does the gas fill the whole box?

1.2.2 Classification of Matter

Matter in bulk can be sorted with three questions (Fig. 1.2). The first: can a physical method such as sieving, filtering, a magnet, evaporation or distillation separate it into simpler parts? If yes, it is a mixture (मिश्रण); if no, it is a pure substance.

4/4Breaks down in a reaction: compound; cannot be broken down: element.

Fig. 1.2Three yes-or-no questions sort any sample of matter into four kinds.NCERT Fig. 1.2 · p. 5

A decision chart that grows step by step. Step 1: a box labelled Matter with the question 'Physically separable?'. Step 2: yes leads to Mixture, no to Pure substance. Step 3: under Mixture, 'Same all through?' leads to Homogeneous (brine, air) or Heterogeneous (sand in water), each with a particle sketch. Step 4: under Pure substance, 'Breaks down chemically?' leads to Compound (water, salt) or Element (copper, oxygen).

Definition
Mixture

Two or more pure substances mixed in any proportion, each keeping its own properties; separable by physical methods.

ExampleBrass, sea water, muddy water.

The substances in a mixture are its components, and their proportions are up to you: nimbu-paani can be made sweeter or sourer at will. If the components are spread evenly down to the level of particles, so that a sip from the top tastes like a sip from the bottom, the mixture is homogeneous, like brass, vinegar or well-stirred nimbu-paani. If the make-up changes from one part to another, often with visible bits, it is heterogeneous, like muddy water, fruit salad or iron filings stirred into sulfur powder. Physical methods pull mixtures apart: a magnet lifts out the iron filings, a filter holds back the mud and evaporation leaves salt behind from sea water.

A pure substance has a fixed composition and the same properties in every sample, like distilled water, copper, sucrose or baking soda. Sucrose from sugarcane and from sugar beet holds carbon, hydrogen and oxygen in identical proportions by mass, and no filtering or evaporating will pull those elements apart.

Pure substances are of two kinds. An element (तत्व) contains only one kind of atom. Its atoms may stay single, as in helium or a piece of iron, or join into molecules, as in two-atom chlorine, three-atom ozone or eight-atom sulfur (Fig. 1.3). A compound (यौगिक) is made when different elements join chemically, always in the same proportion by mass; pulling them apart again takes a chemical reaction.

5/5One kind of atom, fixed combinations, or a jumble of substances.

Fig. 1.3Elements have one kind of atom, single or in molecules; a compound has identical molecules of different atoms; a mixture holds different substances side by side.NCERT Fig. 1.3, 1.4 · p. 6

Four particle panels with a key (N medium circle, O smaller circle, H small circle, Ar large dashed circle). Step 1 highlights 'Element: argon', identical single atoms. Step 2 highlights 'Element: oxygen', identical two-atom molecules. Step 3 highlights 'Compound: ammonia', identical molecules of one nitrogen and three hydrogen atoms. Step 4 highlights 'Mixture: air', nitrogen pairs, an oxygen pair and an argon atom jumbled together. Step 5 shows all four.

Definition
Element

A pure substance made of only one kind of atom; chemical reactions cannot break it into simpler substances.

ExampleCopper (Cu), oxygen (OX2), sulfur (SX8).

Definition
Compound

A pure substance of two or more elements joined chemically in a fixed ratio by mass, with properties unlike those of its elements.

ExampleWater (HX2O), ammonia (NHX3).

A compound is a genuinely new substance. Sodium is a soft metal that reacts violently with water and chlorine is a poisonous green gas, yet together they form table salt. A compound's properties are not an average of its elements' properties.

Quick check
1 of 3
Which of these is a homogeneous mixture?
Show answer

Answer (B) Brass

Brass is an alloy with copper and zinc atoms spread evenly. Ice in water is one substance in two states, distilled water is pure, and sand in water is heterogeneous.

1.3 Properties of Matter and their Measurement

Every substance has a set of properties that identify it, much as a fingerprint identifies a person. To use them in science we must observe and, wherever possible, measure them.

1.3.1 Physical and Chemical Properties

Melting ice still leaves you with water, so a melting point is a physical property: you can observe it while the substance keeps its identity. Burning turns paper into ash and gases, so flammability, which shows itself only as the substance changes into new ones, is a chemical property. A silver anklet has both kinds: its shine and density are physical, while its slow blackening in air, as silver reacts with traces of sulfur compounds, is chemical.

Compare

Two kinds of property

Physical

  • Observed without changing what the substance is
  • Colour, smell, melting and boiling points, density, solubility, hardness

Chemical

  • Observed only through a chemical change
  • Burning, reaction with acids or bases, rusting, tarnishing

Chemists use both kinds to describe, explain and predict how substances behave. The rest of this section is about measuring physical properties.

1.3.2 Measurement of Physical Properties

Saying a door is '2.1 tall' tells nobody anything; '2.1 m' tells everyone. A measured quantity is always a number together with a unit, the number saying how many of that unit fit into it.

Your grandparents may remember buying ghee by the seer and gold by the tola; India began switching to metric units in 1958. The metric system, created in France in the 1790s, steps up and down in powers of ten, so conversions are easy, unlike Britain's English system of inches, feet and pounds. Science needed more than that: one system agreed by every country.

1.3.3 The International System of Units (SI)

Seven base quantities, each with one base unit, form the core of the International System of Units, or SI after its French name, Le Système International d'Unités; speed, volume, density and every other quantity are derived from them. SI was adopted in 1960 at the 11th meeting of the General Conference on Weights and Measures (CGPM), the body behind the Metre Convention, a treaty signed in Paris in 1875.

Table 1.2The seven SI base quantitiesNCERT Table 1.1 · p. 7
Base quantitySymbolSI unitUnit symbolChemists meet it in
lengthlmetrembond lengths
massmkilogramkgweighing reactants
timetsecondsreaction rates
electric currentIampereAelectrolysis
thermodynamic temperatureTkelvinKheating and cooling
amount of substancenmolemolcounting particles
luminous intensityIvcandelacdlight sources

Since 20 May 2019 every base unit is fixed by giving an exact value to a constant of nature, so the unit can be reproduced in any well-equipped laboratory: the second through a frequency of the caesium-133 atom, the metre through the speed of light, the kilogram through the Planck constant, and so on.

Table 1.3The constant behind each base unit (exact since 2019)NCERT Table 1.2 · p. 8
UnitFixed constantExact value
secondcaesium frequency ΔνCs9 192 631 770 Hz
metrespeed of light c299 792 458 m s−1
kilogramPlanck constant h6.62607015×10−34 J s
ampereelementary charge e1.602176634×10−19 C
kelvinBoltzmann constant k1.380649×10−23 J K−1
moleAvogadro constant NA6.02214076×1023 mol−1
candelaluminous efficacy Kcd683 lm W−1

Prefixes scale SI units by powers of ten, so we avoid strings of zeros: a carbon–carbon bond is about 154 pm long and a tablet may hold 500 mg of medicine.

Table 1.4Common SI prefixesNCERT Table 1.3 · p. 9
PrefixSymbolFactorExample
femtof10−15a bond vibrates in ~10 fs
picop10−12atomic radii ~100 pm
nanon10−9visible light 400–700 nm
microµ10−6a hair ~70 μm wide
millim10−3a 500 mg tablet
centic10−2a 15 cm test tube
decid10−11 dm3=1 L
kilok103a 14.2 kg LPG cylinder
megaM106a 2 MW wind turbine
gigaG109a 1 GHz processor

The full list runs from yocto (10−24) to yotta (1024), and since 2022 from quecto (10−30) to quetta (1030).

1.3.4 Mass and Weight

Carry a 1.00 kg packet of atta to the Moon and it holds exactly the same flour: its mass, the amount of matter in it, has not changed. Yet it presses on your hand only a sixth as hard, because its weight, the pull of gravity on it, depends on where it is: about 9.8 N in Delhi, 1.6 N on the Moon. In the lab, mass is found with an analytical balance that reads to 0.0001 g inside a glass case that keeps draughts off the pan. The SI unit is the kilogram, but lab amounts are small, so chemists mostly use grams (1 kg=1000 g).

1.3.5 Volume

A school water tank holds about a cubic metre; a beaker, a few hundred cubic centimetres. Both numbers describe volume, the room a substance takes up, measured in length cubed. The SI unit, m3, is far too big for bench work, so chemists use dm3 and cm3, and the litre (L), which is not an SI unit but is accepted alongside it.

Volume units
1 m3=1000 L1 L=1 dm31 dm3=1000 cm31 cm3=1 mL
(1.1)

3/3A 1 cm cube is 1 mL; a thousand fill 1 L.

Fig. 1.4Each step down divides the edge by 10 and the volume by 1000.NCERT Fig. 1.6 · p. 9

Step 1: a large cube with 1 m edges, labelled 1 cubic metre equals 1000 litres. Step 2: a ten-by-ten grid on its face and one corner cube linked to a medium cube with 10 cm edges, labelled 1 cubic decimetre equals 1 litre. Step 3: a grid on the medium cube and one corner cube linked to a tiny cube with 1 cm edges, labelled 1 cubic centimetre equals 1 millilitre.

Table 1.5Measuring volumes in the laboratoryNCERT Fig. 1.7 · p. 10
DeviceBest used forTypical reading
measuring cylinderquick, rough volumesnearest 1 mL (100 mL size)
burettea variable volume, drop by dropmarked every 0.1 mL
pipetteone fixed volume, exactlye.g. 25.00 mL
volumetric flaskmaking up a solutione.g. 250.0 mL at the mark

1.3.6 Density

Why does a steel spoon sink while a wooden one floats? Steel packs far more mass into each cubic centimetre: it has a higher density, the mass per unit volume:

Density
ρ=mV
(1.2)
where
ρ
density
m
mass
V
volume

Chemists usually quote g cm−3 (the same as g mL−1) rather than the SI kg m−3; 1 g cm−3=1000 kg m−3. Water is 1.00, iron 7.87 and gold 19.3 g cm−3: a one-litre carton full of gold would weigh about 19 kg. Ice (0.92 g cm−3) has an open molecular framework, which is why it floats.

1.3.7 Temperature

A fever of 102 °F on a home thermometer, 38.9 °C on a hospital chart and 312 K in a physics lab are the same temperature on three scales. The kelvin (K) is the SI unit. The Celsius scale puts the freezing and boiling points of water at normal pressure at 0 °C and 100 °C; the Fahrenheit scale puts them at 32 °F and 212 °F (Fig. 1.5).

Celsius to Fahrenheit
tF=95 tC+32
(1.3)
where
tC
temperature in °C
tF
temperature in °F
Celsius to kelvin
T=tC+273.15
(1.4)
where
T
temperature in K

A kelvin and a Celsius degree are the same size; the scales differ only in where zero is. The zero of the kelvin scale, 0 K=−273.15 ∘C, is absolute zero, the lowest temperature possible, so kelvin temperatures are never negative.

2/4Freezing to boiling is 100 K, 100 °C but 180 °F: hence the factor 9/5.

Fig. 1.5The same temperatures on three scales; a Fahrenheit degree is only 5/9 of a kelvin.NCERT Fig. 1.8 · p. 10

Three number lines, for K, degrees Celsius and degrees Fahrenheit, drawn to scale. Step 1 marks ice melting: 273, 0 and 32. Step 2 adds water boiling: 373, 100 and 212, with brackets showing that the gap is 100 K, 100 degrees Celsius but 180 degrees Fahrenheit. Step 3 adds absolute zero: 0 K, minus 273 and minus 460. Step 4 adds 233 K, where Celsius and Fahrenheit both read minus 40.

Example

A Delhi summer on three scales

A Delhi weather station records 46.8 ∘C one May afternoon. (a) Express this in °F and in K. (b) At what temperature do the Celsius and Fahrenheit scales agree?

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

Quick check
1 of 3
Which of these is a chemical property of magnesium?
Show answer

Answer (D) It burns in air to form a white powder

Burning turns magnesium into magnesium oxide, a new substance. Colour, density and melting point can be found without changing the metal.

1.4 Uncertainty in Measurement

Every number in chemistry came from an instrument or from a calculation built on such numbers. Three habits keep them manageable and honest: a compact way to write huge and tiny numbers, a way to show how certain a measurement is, and a safe way to change units.

1.4.1 Scientific Notation

Atoms and molecules are extremely light and extremely numerous. A single drop of water (0.05 g) holds about 1 670 000 000 000 000 000 000 molecules, and a single carbon atom has a mass of about 0.000 000 000 000 000 000 000 019 9 grams. Handling such numbers digit by digit invites mistakes.

Scientific notation writes a number as N×10n, where the digit term N satisfies 1≤N<10 and the exponent n is a whole number. The two numbers above become 1.67×1021 and 1.99×10−23 g. Moving the decimal point k places left adds k to the exponent; moving it right subtracts k: 45 600=4.56×104 and 0.000 093=9.3×10−5.

  • Multiplying or dividing: multiply or divide the digit terms and add or subtract the exponents; then adjust so that 1≤N<10.
  • Adding or subtracting: first rewrite the numbers with the same exponent, then add or subtract the digit terms.
Example

Arithmetic in scientific notation

Evaluate (a) (4.8×106)×(3.5×10−3), (b) (7.2×105)÷(2.4×10−3) and (c) (5.1×10−3)−(8.7×10−4).

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

1.4.2 Significant Figures

No measurement is perfect: the instrument has a limited scale and the person reading it limited skill. A kitchen scale might show a bag of dal as 0.50 kg, while a laboratory balance gives more digits for the same bag. The digits we write show how much we trust a number.

The significant figures (सार्थक अंक) of a measurement are all the digits known for certain plus one final, estimated digit. Writing 24.36 mL says that 24.3 is certain and the 6 is a best estimate. Unless told otherwise, readers take that last digit to be uncertain by one unit either way. Fig. 1.6 shows where the estimated digit comes from.

4/4Write 12.36 mL: four significant figures, the last one estimated.

Fig. 1.6Read the bottom of the meniscus at eye level. The marks give the certain digits; the last digit is an estimate between marks.NCERT p. 12

A magnified burette scale from 12 to 13 mL in steps of 0.1 mL, with the curved liquid surface and a dashed eye-level line. Step 1 shows the scale. Step 2 shades the gap between 12.3 and 12.4 and labels 'certain: 12.3'. Step 3 points at the meniscus and labels 'estimated: 6'. Step 4 writes 12.36 mL, underlines 1, 2 and 3 as certain and the 6 with a dashed line as estimated, and states 4 significant figures.

To count significant figures in a number someone else wrote:

  1. Every non-zero digit counts: 4.71 has three.
  2. Zeros before the first non-zero digit only place the decimal point and do not count: 0.0082 has two.
  3. Zeros between non-zero digits count: 30.05 has four.
  4. Zeros at the end count if there is a decimal point: 2.500 has four, 70.0 has three. Without one they are not counted: 7000 has one. Use scientific notation to show more: 7.0×103 has two, 7.00×103 three.
  5. Exact numbers, such as counts (11 players in a team) and defined relations (1 m=100 cm), have unlimited significant figures.
  6. In scientific notation every digit of N counts: 6.020×1023 has four.
InteractiveSignificant-figure checker

A text box for a number such as 0.004050 or 7000. Each digit appears in a tile, marked significant (solid underline and tick) or not (grey dashed outline), with a sentence on the rule for each zero. Below, the number appears in scientific notation with its count of significant figures, and buttons round it to one to four significant figures, an exact 5 going to the even neighbour. Example buttons load tricky cases.

Fig. 1.7Type a measured number. The checker marks each digit, names the rule, writes the number in scientific notation and rounds it as this section describes.

Careful digits are not enough; results must also be trustworthy. Precision (परिशुद्धता) is how closely repeated measurements agree with each other. Accuracy (यथार्थता) is how close a result is to the true value. A balance that was never zeroed can give beautifully precise readings that are all wrong by the same amount (Fig. 1.8).

4/4Bunched on the true value: precise and accurate.

Fig. 1.8Three students weigh the same 5.00 g standard four times. Bunching shows precision; closeness to the dashed line shows accuracy.NCERT Table 1.4 · p. 13

A number line from 4.80 to 5.20 g with a dashed line at the true value, 5.00 g. Step 1 shows only that. Step 2 adds Meera's readings as circles, bunched around 4.86 g: precise, not accurate. Step 3 adds Arjun's as squares, scattered from 4.92 to 5.11 g: not precise, though their mean is near the true value. Step 4 adds Kabir's as diamonds, bunched between 4.99 and 5.01 g: precise and accurate.

A calculated answer can be no more certain than the least certain measurement in it:

  • Adding or subtracting: keep as many decimal places as the measurement with the fewest decimal places.
  • Multiplying or dividing: keep as many significant figures as the measurement with the fewest.

To round, look at the first digit you drop. More than 5: raise the last kept digit (3.476→3.48). Less than 5: leave it (2.843→2.84). Exactly 5 with only zeros after it: make the last kept digit even (7.45→7.4, 7.55→7.6), so rounding errors do not pile up in one direction. If non-zero digits follow the 5, as in 2.451, round up (2.5).

ExampleNCERT p. 13

Reporting a calculated result

Report each result correctly: (a) 24.6 g+3.218 g+0.0952 g; (b) 4.56 cm×1.4 cm; (c) the mean mass of 7 identical coins weighing 15.0 g together.

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

1.4.3 Dimensional Analysis

The factor-label method (also called the unit factor method or dimensional analysis) changes units safely. Start from a relation such as 1 yd=0.9144 m (exact). Dividing one side by the other gives a unit factor, a fraction equal to one:

Unit factors
0.9144 m1 yd=11 yd0.9144 m=1
(1.5)

Multiplying by 1 never changes a quantity. Pick the factor with the unwanted unit in the denominator, so it cancels, and the wanted unit on top. Units cancel, multiply and can be squared or cubed like algebraic symbols, and several factors can be chained in one line.

ExampleNCERT Examples in 1.4.3 · p. 14

Three conversions with unit factors

Convert (a) the 22 yd length of a cricket pitch into metres, (b) the density of mercury, 13.6 g cm−3, into kg m−3 and (c) a highway speed of 72 km h−1 into m s−1.

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

Quick check
1 of 3
How many significant figures are there in 0.04060 g?
Show answer

Answer 4

Leading zeros only place the decimal point. The zero between 4 and 6 counts, and so does the final zero after the decimal point: four.

1.5 Laws of Chemical Combinations

In the late eighteenth century chemists began to weigh what went into and came out of reactions, and to measure gas volumes. Five regularities emerged, and together they became the evidence for atoms.

1.5.1 Law of Conservation of Mass

The French chemist Antoine Lavoisier (1743–1794) weighed everything. Heating tin in sealed glass vessels, he found that the total mass did not change until the vessel was opened and air rushed in. From many such experiments he concluded that reacting substances never gain or lose matter overall, and set the idea out in his textbook of 1789.

LawLaw of conservation of mass

In any physical or chemical change, the total mass of the substances present after the change equals the total mass before it.

The law can look false at first: a burning candle gets lighter and a rusting nail heavier. In both cases a gas is lost or gained: carbon dioxide and water vapour escape from the flame, and the iron takes up oxygen from the air. Weigh the system with those gases and the total stays the same (Fig. 1.9).

2/3The soda reacts and carbon dioxide fizzes out of the vinegar, but the cap traps it: still 152.36 g.

Fig. 1.9While the bottle is closed the mass cannot change; it falls only when carbon dioxide is allowed to escape.NCERT p. 14

A capped plastic bottle of vinegar, with a small tube of baking soda standing inside, on a digital balance. Step 1, closed, before mixing: 152.36 g. Step 2, still closed: the tube has tipped, bubbles fill the liquid, and the reading is still 152.36 g. Step 3, cap off: dashed arrows show gas leaving the bottle and the reading has fallen to 151.88 g.

1.5.2 Law of Definite Proportions

The French chemist Joseph Proust (1754–1826), working in Spain around 1799, analysed samples of one compound from very different sources and always found the same proportions of elements by mass. Pure water is a modern example.

Table 1.8Composition of pure water from three sourcesNCERT p. 15
Source of the sampleHydrogen (% by mass)Oxygen (% by mass)
Distilled rainwater11.1988.81
Distilled Ganga water11.1988.81
Made by burning hydrogen11.1988.81
LawLaw of definite proportions

A given compound always contains the same elements combined in the same proportions by mass, whatever its source or method of preparation.

Also called the law of definite composition, it lets a chemist identify a substance or check its purity from an analysis. (A few solids, such as iron(II) oxide, closer to FeX0.95O, have slightly variable compositions; you will meet such exceptions later.)

1.5.3 Law of Multiple Proportions

Some pairs of elements form more than one compound, and in 1803 John Dalton noticed a pattern in such families. Carbon and oxygen form carbon monoxide and carbon dioxide: 12.0 g of carbon combines with 16.0 g of oxygen in the first and with 32.0 g in the second. For the same mass of carbon, the masses of oxygen are in the ratio 16:32=1:2, a ratio of small whole numbers.

LawLaw of multiple proportions

If two elements can form more than one compound, then for a fixed mass of the first element, the masses of the second found in the different compounds form a ratio of small whole numbers.

Table 1.9Three oxides of nitrogen
CompoundMass of oxygen per 14.0 g of nitrogenRatio
dinitrogen oxide, NX2O8.0 g1
nitrogen monoxide, NO16.0 g2
nitrogen dioxide, NOX232.0 g4

1.5.4 Gay Lussac's Law of Gaseous Volumes

In 1808 the French chemist Joseph Louis Gay-Lussac (1778–1850) measured the volumes of reacting gases. One volume of hydrogen reacts with exactly one volume of chlorine to give two volumes of hydrogen chloride; one volume of nitrogen and three of hydrogen give two of ammonia, all measured at the same temperature and pressure.

LawGay Lussac's law of gaseous volumes

Measured at one temperature and pressure, the volumes of gases that react, and of any gaseous products, stand in simple whole-number ratios.

In effect it fixes proportions for gas volumes, and it posed a puzzle that took half a century to settle.

1.5.5 Avogadro's Law

In 1811 the Italian scientist Amedeo Avogadro (1776–1856) suggested that a given volume of any gas holds the same number of molecules as the same volume of any other, if temperature and pressure match. His key move was to separate the atom from the molecule: a molecule of an element could contain several atoms.

LawAvogadro's law

At a given temperature and pressure, a fixed volume of any gas contains the same number of molecules, whatever the gas.

Fig. 1.10 shows why this mattered. If hydrogen and chlorine were single, indivisible atoms, one volume of each would hold equal numbers of atoms, and pairing them could give only one volume of hydrogen chloride; Gay-Lussac measured two. If each molecule holds two atoms, every HX2 plus ClX2 yields two HCl molecules, and the volume doubles as observed.

3/3With HX2 and ClX2 molecules, each pair gives two HCl molecules, so the volume doubles: right.

Fig. 1.10Only diatomic molecules explain why one volume of hydrogen and one of chlorine give two volumes of hydrogen chloride.NCERT Fig. 1.9 · p. 16

Two gas syringes of one volume each (hydrogen, chlorine) above a syringe twice as long holding two volumes of hydrogen chloride; a key shows hydrogen as small and chlorine as large circles. Step 1 shows only the volumes. Step 2 fills them with single atoms: four H and four Cl give four HCl filling half the product syringe, the rest marked 'predicted: empty', verdict 'Single atoms: 1 volume, wrong'. Step 3 uses diatomic molecules: four H2 and four Cl2 give eight HCl filling the whole syringe, verdict '2 volumes, correct'.

Dalton and most chemists of his day would not accept that identical atoms could pair up, so Avogadro's idea was largely ignored. In 1860 chemists met in Karlsruhe, Germany, at the first international chemistry conference, to settle the confusion over atomic masses. There the Italian chemist Stanislao Cannizzaro handed out a pamphlet showing that Avogadro's hypothesis gave consistent atomic masses, and the idea was finally accepted.

Quick check
1 of 3
One oxide of sulfur contains 32 g of sulfur combined with 32 g of oxygen; another contains 32 g of sulfur with 48 g of oxygen. Which law do these data illustrate?
Show answer

Answer (C) Law of multiple proportions

Two compounds of the same two elements: for a fixed 32 g of sulfur the oxygen masses are 32:48=2:3, a small whole-number ratio. That is multiple proportions; definite proportions would talk about only one compound.

1.6 Dalton's Atomic Theory

The Greek philosopher Democritus (about 460–370 BCE) called the smallest particles of matter atomos, 'uncuttable', and Kanada's paramanu were a similar idea, but both were philosophical guesses. The English schoolteacher John Dalton (1766–1844) made atoms scientific by linking them to the masses in the laws above. His A New System of Chemical Philosophy (1808) claimed, in modern words:

  1. All matter is made of extremely small particles called atoms, which cannot be divided or destroyed in chemical changes.
  2. All atoms of a given element are identical, including in mass; atoms of different elements have different masses.
  3. Compounds form when atoms of different elements join in fixed, small whole-number ratios.
  4. A chemical reaction only rearranges atoms. No atom is created or destroyed.

These claims explain the mass laws. If atoms are only rearranged (claim 4), total mass cannot change (Fig. 1.11). A fixed whole-number ratio of atoms of fixed mass (claims 2 and 3) gives a fixed mass composition. And since atoms combine as whole units, a second compound of the same elements has whole extra atoms, hence the small whole-number ratios of multiple proportions.

3/3New bonds form: one carbon dioxide and two water molecules, built from exactly the same atoms.

Fig. 1.11In a reaction the atoms only change partners: one carbon, four hydrogen and four oxygen atoms before and after.NCERT p. 16

Step 1: the reactants, a methane molecule (one carbon with four hydrogens) and two oxygen molecules, counted 'C 1, H 4, O 4'. Step 2: all nine atoms pulled apart into a central row with the same count. Step 3: the same atoms regrouped as one linear carbon dioxide molecule and two bent water molecules, with the same count. An arrow is labelled 'atoms regroup'.

The theory had gaps. It could not explain Gay-Lussac's volume ratios, since Dalton rejected molecules of two identical atoms, and it said nothing about why atoms combine. Later work showed that atoms can be split into electrons, protons and neutrons (Unit 2) and that atoms of one element may differ in mass (isotopes, Section 1.7.2). Still, atoms rearranging in fixed ratios is the basis of every chemical equation.

Quick check
1 of 2
Which of Dalton's claims best explains the law of multiple proportions?
Show answer

Answer (B) Atoms combine as whole units in small whole-number ratios

A second compound of the same elements has whole extra atoms (CO versus COX2), so the extra oxygen mass is a whole multiple. Conservation of atoms explains conservation of mass instead.

1.7 Atomic and Molecular Masses

If reactions rearrange atoms in fixed ratios, knowing how heavy each kind of atom is turns ratios of atoms into ratios of grams.

1.7.1 Atomic Mass

Atomic masses are relative masses. Since 1961 the reference has been the carbon-12 atom (12C), assigned exactly 12 units. Why relative? A carbon atom, about 2×10−23 g, is far too light for any balance, so nineteenth-century chemists could only compare atoms through the masses of elements that combine, starting by calling hydrogen 1. A modern mass spectrometer measures atomic masses directly and precisely, but still reports them in these units.

Definition
Atomic mass unit (u)

Exactly 1/12 of the mass of a single atom of carbon-12, that is 1.6605×10−24 g; formerly written amu, also called the dalton (Da).

Atomic mass unit
1 u=112 m ⁣(12C)=1.6605×10−24 g
(1.6)

To express an atom's mass in u, divide its mass in grams by 1.6605×10−24 g: a chlorine-35 atom of 5.8067×10−23 g has an atomic mass of 34.969 u. The old 'amu' is now written u, the unified atomic mass unit.

1.7.2 Average Atomic Mass

Most natural elements are mixtures of isotopes, atoms of one element with different masses. The atomic mass in the periodic table is the average over a natural sample, each isotope weighted by how common it is:

Average atomic mass
Aˉ=∑ixiAi
(1.7)
where
xi
fractional abundance of isotope i (per cent ÷ 100)
Ai
atomic mass of isotope i

3/3Weighting each mass by its abundance gives 63.55 u, the value in the periodic table.

Fig. 1.12The weighted average sits where the bars would balance, nearer the more abundant isotope.NCERT p. 17

A mass axis from 62 to 66 u. Step 1: a tall bar for copper-63 at 62.93 u (69.15 per cent) and a shorter dashed bar for copper-65 at 64.93 u (30.85 per cent). Step 2: a dashed line at the plain average, 63.93. Step 3: that line fades and a solid line with a balance-point triangle marks the weighted average, 63.55 u, nearer the copper-63 bar.

ExampleNCERT p. 17

Average atomic mass of copper

Natural copper contains 69.15 % of 63Cu (62.9296 u) and 30.85 % of 65Cu (64.9278 u). Find the average atomic mass of copper.

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

1.7.3 Molecular Mass

The molecular mass is the sum of the atomic masses of all the atoms in one molecule. For ammonia, NHX3: 14.007+3(1.008)=17.031 u. For carbon dioxide, COX2: 12.011+2(15.999)=44.009 u.

1.7.4 Formula Mass

Ionic compounds have no separate molecules. Solid sodium chloride is a vast lattice of NaX+ and ClX− ions, each surrounded by six of the other kind, with no 'NaCl molecule' to weigh. We add up the atomic masses of the simplest formula, the formula unit, to get the formula mass. For calcium chloride, CaClX2: 40.078+2(35.45)=110.98 u.

Example 1.1NCERT Problem 1.1 · p. 18

Molecular mass of paracetamol

Paracetamol, the common fever medicine, has the formula CX8HX9NOX2. Calculate its molecular mass. (Atomic masses: C 12.011, H 1.008, N 14.007, O 15.999 u.)

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

Quick check
1 of 3
Find the formula mass of potassium sulfate, KX2SOX4 (K 39.10, S 32.06, O 16.00 u).
u
Show answer

Answer 174.26 u

2(39.10)+32.06+4(16.00)=78.20+32.06+64.00=174.26 u.

1.8 Mole Concept and Molar Masses

A bank weighs a bag of coins instead of counting it, and a paper shop sells sheets by the ream of 500. Chemists face the same problem on a gigantic scale: a pinch of salt holds about 1022 ions. Their counting unit is the mole (मोल); the amount of substance n, in moles, is an SI base quantity.

Definition
Mole (mol)

The SI base unit for amount of substance: a mole of anything is exactly 6.02214076×1023 of its specified entities, whether atoms, molecules, ions, electrons or defined groups of particles.

Example1 mol of NaCl is 6.022×1023 formula units.

The number of entities per mole is the Avogadro constant, NA=6.02214076×1023 mol−1 (the bare number is the Avogadro number), exact by definition since 2019. Its value makes the mole and the u fit together exactly: NA atomic mass units add up to one gram.

Avogadro constant and the u
NA×1 u=6.0221×1023×1.6605×10−24 g=1.000 g
(1.8)

Counting one particle per second, day and night, one mole would take about 1.9×1016 years, over a million times the age of the universe. A mole of hydrogen atoms, of water molecules or of NaCl formula units all contain the same 6.022×1023 entities.

A substance's molar mass (मोलर द्रव्यमान) M is how many grams one mole of it weighs, in g mol−1. Because of Eq. (1.8), M has the same number as the atomic, molecular or formula mass in u: COX2 44.01, NHX3 17.03 and CaClX2 110.98 g mol−1. (Since 2019 the molar mass of carbon-12 is not exactly 12 g mol−1, but it is off by less than a part per billion.)

Definition
Molar mass

The number of grams that one mole of a substance weighs; numerically equal to its atomic, molecular or formula mass in u.

ExampleM(HX2O)=18.02 g mol−1.

Three relations link moles to measurable things; together they form the 'mole map' of Fig. 1.13.

Moles from mass
n=mM
(1.9)
where
m
mass in g
M
molar mass in g mol−1
Number of particles
N=n NA
(1.10)
Volume of a gas at STP
V=n×22.7 L mol−1
(1.11)
where
V
volume of an ideal gas at STP (273.15 K, 1 bar)

4/4For a gas at STP, one mole occupies 22.7 L.

Fig. 1.13Every conversion goes through moles: divide or multiply by the molar mass, the Avogadro constant or the molar volume.NCERT Fig. 1.11 · p. 18

A central box labelled moles. Step 1: it alone, noting 1 mol = 6.022 times ten to the 23 entities. Step 2: a mass box on the left, arrows labelled divide by M (to moles) and multiply by M (back). Step 3: a particles box on the right, arrows multiply by N A and divide by N A. Step 4: a gas volume box below for STP, arrows multiply by 22.7 L and divide by 22.7 L.

Example

Counting molecules in a tablet

A tablet contains 500 mg of paracetamol (M=151.17 g mol−1). How many (a) moles of paracetamol, (b) paracetamol molecules and (c) carbon atoms does it contain?

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

Quick check
1 of 3
How many water molecules are there in 9.0 g of water (M=18.0 g mol−1)? Give your answer in scientific notation.
Show answer

Answer 3.011e+23

n=9.0/18.0=0.50 mol; N=0.50×6.022×1023=3.0×1023 molecules.

1.9 Percentage Composition

For a new compound the first question is which elements it contains, and in what proportions; for a known one, the same data check its purity, as when a fertiliser's nitrogen content is tested against its label. The mass percentage of an element follows from the formula:

Mass per cent of an element
wE=a AEM×100 %
(1.12)
where
a
number of atoms of element E in the formula
AE
atomic (molar) mass of E
M
molar mass of the compound
Example

Which fertiliser carries more nitrogen?

Compare the mass percentage of nitrogen in urea, CO(NHX2)X2, and in ammonium sulfate, (NHX4)X2SOX4. (C 12.011, H 1.008, N 14.007, O 15.999, S 32.06 u.)

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

1.9.1 Empirical Formula and Molecular Formula

Percentages also work backwards, giving the formula. The empirical formula (मूलानुपाती सूत्र) shows the simplest whole-number ratio of atoms; the molecular formula shows the actual numbers in one molecule. Glucose, CX6HX12OX6, has the empirical formula CHX2O, benzene (CX6HX6) has CH, and for water both formulas are HX2O.

Definition
Empirical formula

The formula that shows the simplest whole-number ratio of the atoms of each element in a compound.

ExampleCHX2O for glucose, CX6HX12OX6.

Mass percentages give the empirical formula; the molar mass then gives the molecular formula (Fig. 1.14):

  1. Assume a 100 g sample, so each mass percentage becomes a mass in grams.
  2. Divide each mass by the element's atomic mass to get moles of atoms.
  3. Divide every mole value by the smallest one.
  4. If the ratios are not whole numbers, multiply all of them by a small factor: a ratio ending in .5 needs ×2; one ending in .33 or .67 needs ×3. Never round 1.5 to 2.
  5. Write the empirical formula and add up its empirical formula mass.
  6. Find n=M/(empirical formula mass), which must be a whole number, and multiply every subscript by n.

4/4n=62.07/31.03=2, so the molecular formula is CX2HX6OX2.

Fig. 1.14Grams become moles, moles become a ratio, and the molar mass scales the ratio up to the real formula.NCERT p. 20

Step 1: a 100 g bar split by mass into carbon 38.7 g, a thin hydrogen slice of 9.7 g and oxygen 51.6 g. Step 2: a bar of moles, carbon 3.22, hydrogen 9.66 and oxygen 3.22, so hydrogen now has the widest share. Step 3: dividing by the smallest gives C : H : O = 1 : 3 : 1, the empirical formula CH3O. Step 4: molar mass 62.07 divided by empirical formula mass 31.03 gives n = 2, so the molecular formula is C2H6O2.

Example 1.2NCERT Problem 1.2 · p. 19

The formula of a coolant

Ethylene glycol, used as antifreeze in vehicle radiators, contains 38.70 % carbon, 9.74 % hydrogen and 51.56 % oxygen by mass. Its molar mass is 62.07 g mol−1. Find its empirical and molecular formulas. (C 12.011, H 1.008, O 15.999)

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

Quick check
1 of 3
What is the mass percentage of oxygen in calcium carbonate, CaCOX3 (Ca 40.08, C 12.01, O 16.00)?
%
Show answer

Answer 47.96 %

M=40.08+12.01+48.00=100.09 g mol−1; wO=48.00/100.09×100 %=47.96 %.

1.10 Stoichiometry and Stoichiometric Calculations

Stoichiometry (रससमीकरणमिति), from the Greek stoicheion ('element') and metron ('measure'), uses a balanced equation to calculate the masses, or gas volumes, of reactants needed and products formed. Take the burning of propane, the gas in camping stoves:

Burning propane
CX3HX8(g)+5 OX2(g)→3 COX2(g)+4 HX2O(g)
(1.13)

Propane and oxygen are the reactants, carbon dioxide and water the products. State symbols show (s) solid, (l) liquid, (g) gas or (aq) dissolved in water. The stoichiometric coefficients in front (a missing one means 1) give the ratio in which molecules, and so moles, react and form:

Table 1.10Four readings of Eq. (1.13), with water as vapourNCERT p. 20
CX3HX8OX2COX2HX2O
molecules1534
moles1534
gas volumes (same T, p)1534
mass in g44.1160.0132.072.1

The volume row follows from Avogadro's law and the mass row from molar masses; reactants and products both total 204.1 g, as conservation of mass demands. Because coefficients are mole ratios, every problem follows one path: convert the data to moles, apply the mole ratio, convert back to the units you need.

Step by step

Balancing the burning of butane

Step 1 of 5

Write the formulas

Butane burns in oxygen to give carbon dioxide and water: CX4HX10+OX2→COX2+HX2O (not yet balanced).

1 / 5
Example 1.3NCERT Problem 1.3 · p. 22

Carbon dioxide from a gas cylinder

A domestic LPG cylinder holds 14.2 kg of gas. Treating the gas as pure butane, CX4HX10 (M=58.12 g mol−1), what mass of carbon dioxide (M=44.01 g mol−1) forms when all of it burns?

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

Example 1.4NCERT Problem 1.4 · p. 22

How much magnesium?

How many moles, and how many grams, of magnesium must burn to make 10.0 g of magnesium oxide? (Mg 24.31, O 16.00)

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

1.10.1 Limiting Reagent

Reactants are rarely mixed in exactly the equation's ratio. A factory with 120 auto-rickshaw bodies and 300 wheels can finish only 300/3=100 autos, leaving 20 bodies: the wheels limit the output. In a reaction, the reactant used up first stops it, however much of the others remains.

Definition
Limiting reagent

The reactant completely used up first; it fixes the maximum amount of product, and the other reactants are in excess.

To find the limiting reagent (सीमांत अभिकर्मक), convert each reactant to moles and divide by its coefficient; the smallest result is limiting. Raw masses, or even raw moles, can mislead (Fig. 1.15).

3/3Six HX2O form and two OX2 remain: hydrogen is the limiting reagent.

Fig. 1.15Two hydrogen molecules are needed for every oxygen molecule, so hydrogen runs out first even though there are more hydrogen molecules than oxygen molecules.NCERT p. 21

Before and after boxes joined by an arrow. Step 1: six hydrogen molecules (pairs of small open circles) and five oxygen molecules (pairs of larger shaded circles). Step 2: dashed loops around three groups of two hydrogen and one oxygen molecule; two oxygen molecules are marked spare. Step 3: the after box holds six water molecules and two leftover oxygen molecules, with the caption 'hydrogen runs out first: limiting'.

Example 1.5NCERT Problem 1.5 · p. 22

Welding rails with thermite

Indian Railways welds rails with the thermite reaction, 2 Al+FeX2OX3→AlX2OX3+2 Fe, which gives molten iron. A charge holds 54.0 g of Al and 200 g of FeX2OX3. Find (a) the limiting reagent, (b) the mass of iron formed and (c) the mass of excess reactant left. (Al 26.98, Fe 55.85, FeX2OX3 159.69 g mol−1)

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

InteractiveStoichiometry lab

Buttons choose one of five balanced reactions (hydrogen burning, ammonia synthesis, methane burning, thermite, zinc with hydrochloric acid). Two sliders with number boxes set the reactant masses in grams. A table gives each reactant's molar mass, moles and moles divided by its coefficient, naming the smallest as limiting. Bars show the moles of every substance before and after, and a status line gives the product masses, the excess left over and a check that total mass is unchanged.

Fig. 1.16Pick a reaction and set the mass of each reactant. The lab converts to moles, finds the limiting reagent and shows how much of everything is left at the end.
Quick check
1 of 3
For 2 HX2+OX2→2 HX2O, a vessel holds 3.0 mol of HX2 and 2.0 mol of OX2. Which is the limiting reagent?
Show answer

Answer (B) Hydrogen

HX2: 3.0/2=1.5; OX2: 2.0/1=2.0. Hydrogen is limiting; 0.5 mol of oxygen is left over even though there was less of it.

1.10.2 Reactions in Solutions

Most lab reactions are run in solution, where dissolved particles meet easily, so we need the solution's concentration: how much solute a given amount holds. Four measures are common.

1. Mass per cent (w/w %) is the mass of solute in 100 g of solution:

Mass per cent
w=msolutemsolution×100 %
(1.14)
where
msolution
mass of solute + mass of solvent

2. Mole fraction is the fraction of all the moles in the solution that belong to one component. For a solution of A in B:

Mole fraction
xA=nAnA+nBxA+xB=1
(1.15)

The mole fractions (मोल अंश) of all the components add up to 1, and a mole fraction has no unit.

Example 1.6NCERT Problem 1.6 · p. 23

A urea spray for crops

Farmers sometimes spray crops with a urea solution. One is made by dissolving 25.0 g of urea (M=60.06 g mol−1) in 225 g of water (M=18.02 g mol−1). Find (a) the mass per cent of urea and (b) the mole fractions of urea and water.

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

3. Molarity (M) is the most widely used measure in the laboratory.

Definition
Molarity

The number of moles of solute in one litre of solution; unit mol L−1, written M.

Example0.20 mol of NaCl made up to 1.0 L of solution is 0.20 M.

Molarity
molarity=nV
(1.16)
where
n
moles of solute
V
volume of solution in litres

Solutions of known molarity (मोलरता) are usually made by diluting a more concentrated stock solution. Adding water changes the volume but not the number of moles of solute, so moles before equal moles after (Fig. 1.17):

Dilution
M1V1=M2V2
(1.17)
where
M1,V1
molarity and volume of the stock taken
M2,V2
molarity and volume after dilution

3/3Add water to the mark and mix: still 0.150 mol, now in 0.500 L, so 0.30 M.

Fig. 1.17Dilution spreads the same moles of solute through a bigger volume.NCERT p. 23

Step 1: a beaker labelled stock 6.0 M, crowded with solute dots, and a pipette holding 25.0 mL with six dots. Step 2: the pipette has emptied into a 500 mL volumetric flask; six dots sit in a little liquid, labelled 0.150 mol. Step 3: the flask is filled to the mark; the same six dots spread through it, labelled 0.30 M, with the note 'same moles, 20 times the volume'.

Example

Diluting a stock acid

What volume of 6.0 M hydrochloric acid must be diluted to make 500 mL of 0.30 M acid?

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

Example 1.7NCERT Problem 1.7 · p. 24

Molarity of a glucose drip

A 500 mL hospital drip bag contains 25.0 g of glucose, CX6HX12OX6 (M=180.16 g mol−1). What is the molarity of the glucose?

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

Molarity has a drawback: a warmed solution expands, so the same moles occupy more litres and the molarity falls.

4. Molality (m) avoids this by referring to the mass of the solvent instead of the volume of the solution.

Definition
Molality

The number of moles of solute per kilogram of solvent; unit mol kg−1, written m.

Example0.50 mol of sugar dissolved in 1.00 kg of water is 0.50 m.

Molality
molality=nmsolvent
(1.18)
where
n
moles of solute
msolvent
mass of solvent in kilograms

Masses do not change with temperature, so neither does molality (मोललता). Converting between molarity and molality needs the solution's density.

Example 1.8NCERT Problem 1.8 · p. 24

From molarity to molality

A 2.00 M solution of sulfuric acid, HX2SOX4 (M=98.08 g mol−1), has a density of 1.12 g mL−1. Find its molality.

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

Quick check
1 of 3
A stock solution prepared in a cool storeroom is used in a warm lab. Which measure of its concentration has changed?
Show answer

Answer (D) Molarity

Warming expands the solution, so molarity drops slightly; the other measures use only masses or moles.
Recap

Some Basic Concepts of Chemistry: key points and summary

  1. Chemistry studies what substances are made of, how they are built and how they change; India's chemical tradition runs from Harappan crafts and Kanada's paramanu to modern industry (Development of Chemistry).
  2. Matter has mass and occupies space. Solids, liquids and gases differ in how closely packed and how mobile their particles are, and they interconvert with changes of temperature and pressure (Section 1.2.1).
  3. Mixtures (homogeneous or heterogeneous) can be separated by physical methods; pure substances are elements (one kind of atom) or compounds (elements combined in a fixed ratio) (Section 1.2.2).
  4. A measured quantity is a number with a unit. SI has seven base units, each defined since 2019 by an exact constant of nature, and prefixes from atto to tera and beyond (Section 1.3.3).
  5. Mass is constant but weight varies with gravity; 1 L=1 dm3=1000 mL; density ρ=m/V; T=tC+273.15 (Section 1.3.4, Section 1.3.7).
  6. Scientific notation handles huge and tiny numbers. Significant figures are the certain digits plus one estimated digit; results keep the fewest decimal places (sums) or significant figures (products). Precision is agreement between repeats, accuracy closeness to the true value (Section 1.4.2); unit factors convert units (Section 1.4.3).
  7. Five laws of chemical combination (conservation of mass, definite proportions, multiple proportions, gaseous volumes and Avogadro's law) led to Dalton's atomic theory (Section 1.5, Section 1.6).
  8. Atomic masses are relative to carbon-12 (1 u is one-twelfth of its mass); periodic-table values are abundance-weighted averages of isotopes (Section 1.7).
  9. One mole contains NA=6.02214076×1023 entities; the molar mass in g mol−1 equals the atomic, molecular or formula mass in u; n=m/M (Section 1.8).
  10. Mass percentages give the empirical formula; the molar mass then gives the molecular formula (Section 1.9).
  11. Coefficients in a balanced equation are mole ratios; the limiting reagent, used up first, fixes the amount of product (Section 1.10, Section 1.10.1).
  12. Concentration can be given as mass per cent, mole fraction, molarity (changes with temperature) or molality (does not) (Section 1.10.2).
Practice · 39 questions

Practice questions with answers (NCERT “Exercises”)

  1. NumericalEasyNCERT Exercises 1.1

    Calculate the molar mass of each substance. (H 1.008, C 12.011, N 14.007, O 15.999, S 32.06, Ca 40.078)

    1. Ammonia, NHX3

      g/mol
      Show solution

      14.007+3(1.008)=17.03 g mol−1.

    2. Sulfur dioxide, SOX2

      g/mol
      Show solution

      32.06+2(15.999)=64.06 g mol−1.

    3. Calcium carbonate, CaCOX3

      g/mol
      Show solution

      40.078+12.011+3(15.999)=100.09 g mol−1.

  2. NumericalEasyNCERT Exercises 1.2

    Saltpetre, potassium nitrate (KNOX3), is an ingredient of fireworks. Calculate the mass per cent of each element in it. (K 39.10, N 14.01, O 16.00) Enter the percentage of oxygen.

    %
    Show solution

    M=39.10+14.01+48.00=101.11 g mol−1. K: 39.10/101.11=38.67 %. N: 14.01/101.11=13.86 %. O: 48.00/101.11=47.47 %. Check: the three add up to 100 %.

  3. NumericalMediumNCERT Exercises 1.3

    The black oxide of manganese used in dry cells contains 63.2 % manganese and 36.8 % oxygen by mass. Find its empirical formula. (Mn 54.94, O 16.00)

    Show solution

    Mn: 63.2/54.94=1.150 mol; O: 36.8/16.00=2.300 mol. Divide by 1.150: Mn : O =1:2.00. Empirical formula: MnOX2.

  4. NumericalMediumNCERT Exercises 1.4

    Sulfur burns in oxygen to form sulfur dioxide, S+OX2→SOX2. Calculate the mass of SOX2 (M=64.06 g mol−1) formed in each case. (O 16.00)

    1. 0.50 mol of sulfur burns in plenty of air.

      g
      Show solution

      Oxygen is in excess, so 0.50 mol of SOX2 forms: 0.50×64.06=32.0 g.

    2. 0.50 mol of sulfur burns in 8.0 g of oxygen.

      g
      Show solution

      8.0 g of OX2 is 0.25 mol: oxygen is limiting. 0.25 mol of SOX2 forms, which is 16.0 g.

    3. 1.0 mol of sulfur burns in 48.0 g of oxygen.

      g
      Show solution

      48.0 g of OX2 is 1.5 mol, more than needed: sulfur is limiting. 1.0 mol of SOX2 forms, which is 64.1 g.

  5. NumericalEasyNCERT Exercises 1.5

    What mass of oxalic acid dihydrate, HX2CX2OX4 ⋅ 2 HX2O (M=126.07 g mol−1), is needed to make 250 mL of a 0.0500 M solution for a titration?

    g
    Show solution

    n=0.0500×0.250=0.0125 mol. Mass =0.0125×126.07=1.58 g.

  6. NumericalMediumNCERT Exercises 1.6

    Concentrated hydrochloric acid is 37.0 % HCl by mass and has a density of 1.18 g mL−1. What is its molarity? (HCl 36.46 g mol−1)

    M
    Show solution

    Mass of 1.000 L: 1000×1.18=1180 g. Mass of HCl: 0.370×1180=436.6 g. Moles: 436.6/36.46=11.97 mol, so the acid is about 12.0 M.

  7. NumericalEasyNCERT Exercises 1.7

    How much iron can in principle be obtained from 1.00 kg of pure haematite, FeX2OX3? (Fe 55.85, O 16.00) Give the mass in grams.

    g
    Show solution

    M(FeX2OX3)=2(55.85)+3(16.00)=159.70 g mol−1. Fraction of iron =111.70/159.70=0.6994. Iron =0.6994×1000=699 g.

  8. NumericalMediumNCERT Exercises 1.8

    An oxide of nitrogen contains 30.4 % nitrogen by mass, and its molar mass is 92 g mol−1. Find its molecular formula. (N 14.01, O 16.00)

    Show solution

    N: 30.4/14.01=2.170; O: 69.6/16.00=4.350; ratio 1:2.00. Empirical formula NOX2, with formula mass 46.01. n=92/46.01=2, so the molecular formula is NX2OX4.

  9. NumericalEasyNCERT Exercises 1.9

    Bromine has two natural isotopes: bromine-79 (50.69 %, 78.9183 u) and bromine-81 (49.31 %, 80.9163 u). Calculate the average atomic mass of bromine.

    u
    Show solution

    0.5069(78.9183)+0.4931(80.9163) =40.004+39.900=79.90 u.

  10. NumericalEasyNCERT Exercises 1.10

    A camping-stove canister holds 0.25 mol of propane, CX3HX8. Calculate:

    1. the number of moles of carbon atoms

      mol
      Show solution

      3×0.25=0.75 mol.

    2. the number of moles of hydrogen atoms

      mol
      Show solution

      8×0.25=2.0 mol.

    3. the number of propane molecules

      Show solution

      0.25×6.022×1023=1.51×1023.

  11. NumericalEasyNCERT Exercises 1.11

    9.0 g of glucose (M=180.16 g mol−1) is dissolved in water and the solution is made up to 250 mL. What is its concentration in mol L−1?

    M
    Show solution

    n=9.0/180.16=0.0500 mol. Molarity =0.0500/0.250=0.200 M.

  12. NumericalMediumNCERT Exercises 1.12

    Ethanol (M=46.07 g mol−1) has a density of 0.789 g mL−1. What volume of pure ethanol is needed to make 500 mL of a 0.40 M solution?

    mL
    Show solution

    n=0.40×0.500=0.20 mol. Mass =0.20×46.07=9.21 g. Volume =9.21/0.789=11.7 mL.

  13. NumericalMediumNCERT Exercises 1.13

    Pressure is force per unit area, and 1 Pa=1 N m−2. At a hill station the column of air above each square centimetre of ground has a mass of 820 g. Taking g=9.8 m s−2, find the air pressure there in pascals.

    Pa
    Show solution

    Force =0.820 kg×9.8=8.04 N. Area =1 cm2=1×10−4 m2. Pressure =8.04/1×10−4=8.0×104 Pa, about 80 % of sea-level pressure.

  14. Short answerEasyNCERT Exercises 1.14

    Name the SI unit of mass. How was it defined before 2019, and how is it defined now?

    Show solution

    The kilogram (kg). From 1889 to 2019 it was the mass of a platinum–iridium prototype kept near Paris. Since 2019 it is fixed by the exact Planck constant, h=6.62607015×10−34 J s, with J s=kg m2 s−1 and the metre and second defined through c and the caesium frequency.

  15. Quick answerEasyNCERT Exercises 1.15

    Match each SI prefix with the factor it stands for.

    Tap an item on the left, then its match on the right.

    Show answer
    Answer
    • centi : 10−2
    • nano : 10−9
    • femto : 10−15
    • kilo : 103
    • giga : 109
    centi 10−2, nano 10−9, femto 10−15, kilo 103, giga 109.
  16. Short answerEasyNCERT Exercises 1.16

    What are significant figures? Explain using a burette reading recorded as 18.74 mL.

    Show solution

    They are the meaningful digits of a measurement: the certain digits plus one estimated digit. In 18.74 mL, marks every 0.1 mL make 18.7 certain and the 4 is estimated: four significant figures, uncertain by about ±0.01 mL.

  17. NumericalMediumNCERT Exercises 1.17

    Some groundwater in Rajasthan holds fluoride at 4.5 ppm by mass, three times the WHO guideline of 1.5 mg L−1. (F 19.00)

    1. Express 4.5 ppm as a mass percentage.

      %
      Show solution

      4.5/106×100 %=4.5×10−4 %.

    2. Find the molality of fluoride ions, treating the solvent mass as that of the water.

      mol/kg
      Show solution

      1 kg of water holds 4.5×10−3 g of fluoride, which is 2.4×10−4 mol: about 2.4×10−4 mol kg−1.

  18. Short answerEasyNCERT Exercises 1.18

    Write each number in scientific notation.

    1. 0.000706

      Show solution

      7.06×10−4

    2. 98 100 (measured to three significant figures)

      Show solution

      9.81×104

    3. 700.0

      Show solution

      7.000×102: the zeros after the decimal point are significant.

    4. 3.0060

      Show solution

      3.0060×100; all five digits are kept.

  19. Quick answerEasyNCERT Exercises 1.19

    How many significant figures does each number have?

    1. 0.00610

      Show solution

      Three: 6, 1 and the final 0.

    2. 7007

      Show solution

      Four: zeros between non-zero digits count.

    3. 30.0

      Show solution

      Three: the decimal point makes both zeros significant.

    4. 1.0700

      Show solution

      Five.

  20. Quick answerMediumNCERT Exercises 1.20

    Round each number to three significant figures, using the rounding rules of Section 1.4.2.

    1. 45.362

      Show solution

      Dropped digit 6 > 5: 45.4.

    2. 2.0751

      Show solution

      Dropped part '51' is more than half: 2.08.

    3. 0.006155

      Show solution

      Exactly 5 dropped and the kept digit 5 is odd, so round up: 0.00616.

    4. 7845

      Show solution

      Exactly 5 dropped and the kept digit 4 is even, so it stays: 7.84×103.

  21. Short answerMediumNCERT Exercises 1.21

    Four samples of oxides of sulfur were analysed: (i) 16.0 g S with 16.0 g O; (ii) 16.0 g S with 24.0 g O; (iii) 4.0 g S with 4.0 g O; (iv) 8.0 g S with 12.0 g O.

    1. Which law of chemical combination do the data illustrate? State it.

      Show solution

      Per 16.0 g of S the oxygen masses are 16.0, 24.0, 16.0 and 24.0 g: two compounds (SOX2 and SOX3) with oxygen in the ratio 2:3. That is the law of multiple proportions: for a fixed mass of sulfur, the oxygen masses in different compounds form a small whole-number ratio. (Pairs (i), (iii) and (ii), (iv) also show definite proportions.)

    2. Fill in: 1 m = µm; 1 g = µg; 1 dm3 = mL; 1 cm3 = m3.

      Show answer

      Answer (1) 10^6, (2) 10^6, (3) 1000, (4) 10^-6

      1 m=106 μm; 1 g=106 μg; 1 dm3=1000 mL; 1 cm3=10−6 m3.
  22. NumericalEasyNCERT Exercises 1.22

    A communication satellite in geostationary orbit is about 3.58×104 km above the equator. Taking the speed of light as 3.0×108 m s−1, how long does a radio signal take to reach it from the ground directly below? Give the time in seconds.

    s
    Show solution

    Distance =3.58×107 m. Time =3.58×107/3.0×108=0.12 s.

  23. Quick answerMediumNCERT Exercises 1.23

    For the reaction XX2+3 YX2→2 XYX3, identify the limiting reagent, if any, in each mixture.

    1. 150 molecules of XX2 and 300 molecules of YX2

      Show solution

      150 XX2 would need 450 YX2; only 300 are present, so YX2 is limiting.

    2. 1.0 mol XX2 and 3.0 mol YX2

      Show solution

      1.0/1=3.0/3: the exact ratio, so neither is in excess.

    3. 2.0 mol XX2 and 4.5 mol YX2

      Show solution

      2.0/1=2.0 but 4.5/3=1.5: YX2 is limiting.

    4. 0.50 mol XX2 and 3.0 mol YX2

      Show solution

      0.50/1=0.50 but 3.0/3=1.0: XX2 is limiting.

  24. NumericalHardNCERT Exercises 1.24

    Methanol is made industrially by CO(g)+2 HX2(g) →CHX3OH(l). A reactor is charged with 280 g of CO and 50.0 g of HX2. (C 12.01, O 16.00, H 1.008)

    1. What mass of methanol can form?

      g
      Show solution

      n(CO)=280/28.01=10.0 mol; n(HX2)=50.0/2.016=24.8 mol. CO needs 20.0 mol of HX2, which is available, so CO is limiting. Methanol: 10.0×32.04=320 g.

    2. Which reactant is left over?

      Show solution

      Hydrogen, since only 20.0 of its 24.8 mol react.

    3. What mass of it remains?

      g
      Show solution

      24.80−19.99=4.81 mol, which is 4.81×2.016=9.70 g.

  25. Short answerEasyNCERT Exercises 1.25

    How do '0.20 mol NaCl', '0.20 M NaCl' and '0.20 m NaCl' differ?

    Show solution

    0.20 mol is an amount: 11.7 g of salt, in any volume. 0.20 M is a molarity: 0.20 mol per litre of solution, so 250 mL holds 0.050 mol. 0.20 m is a molality: per kilogram of solvent, independent of temperature.

  26. NumericalEasyNCERT Exercises 1.26

    At a fixed temperature and pressure, 6.0 L of nitrogen reacts with 18.0 L of hydrogen: NX2(g)+3 HX2(g)→2 NHX3(g). What volume of ammonia forms?

    L
    Show solution

    The volumes are in the ratio 1:3, exactly as required, so both react completely. Ammonia: 2×6.0=12.0 L.

  27. Quick answerEasyNCERT Exercises 1.27

    Express each quantity in SI base units (m, kg or s), in scientific notation.

    1. 154 pm, the C–C bond length

      m
      Show solution

      154×10−12 m=1.54×10−10 m.

    2. 3250 mg

      kg
      Show solution

      3250 mg=3.250 g=3.25×10−3 kg.

    3. 75.0 μs

      s
      Show solution

      75.0×10−6 s=7.50×10−5 s.

  28. Quick answerMediumNCERT Exercises 1.28

    Which sample contains the largest number of atoms? (C 12.01, Na 22.99, N 14.01, Ca 40.08)

    Show answer

    Answer (A) 1 g of carbon

    Moles of atoms: C 1/12.01=0.0833; Na 0.0435; N atoms in NX2 1/14.01=0.0714 (diatomic does not matter: 1 g of N is 1 g of N atoms); Ca 0.0250. Carbon has the lightest atoms, so the most atoms.
  29. NumericalHardNCERT Exercises 1.29

    In a methanol–water solution the mole fraction of methanol is 0.050. Estimate its molarity, assuming that the solution has the same volume as the water in it and that water's density is 1.00 g mL−1.

    M
    Show solution

    1000 g of water is 1000/18.02=55.5 mol and (by assumption) about 1 L of solution. x=0.050 gives nMeOH/nwater=0.050/0.950, so nMeOH=55.5×0.0526=2.92 mol. Molarity ≈2.9 M.

  30. NumericalEasyNCERT Exercises 1.30

    What is the mass, in grams, of one water molecule (M=18.02 g mol−1)?

    g
    Show solution

    18.02/6.022×1023=2.99×10−23 g.

  31. Quick answerMediumNCERT Exercises 1.31

    How many significant figures should each answer have?

    1. 0.04115×312.60.0230

      Show solution

      0.0230 has three significant figures, the fewest, so the answer (559) has three.

    2. 4×1.268, where 4 is an exact count

      Show solution

      The exact 4 does not limit the result; 1.268 has four, so 5.072 keeps four.

    3. 0.0217+1.5022+0.104

      Show solution

      The sum is 1.6279; 0.104 has three decimal places, so report 1.628: four significant figures.

  32. NumericalMediumNCERT Exercises 1.32

    Natural magnesium is 78.99 % magnesium-24 (23.98504 g mol−1), 10.00 % magnesium-25 (24.98584 g mol−1) and 11.01 % magnesium-26 (25.98259 g mol−1). Calculate its molar mass.

    g/mol
    Show solution

    0.7899(23.98504)=18.9458 0.1000(24.98584)=2.4986 0.1101(25.98259)=2.8607 Sum: 24.31 g mol−1.

  33. NumericalMediumNCERT Exercises 1.33

    Calculate the number of atoms in each sample. (C 12.01)

    1. 15 mol of neon

      Show solution

      15×6.022×1023=9.03×1024 atoms.

    2. 60 u of carbon-12

      Show solution

      One 12C atom is 12 u, so 60/12=5 atoms.

    3. 60 g of carbon

      Show solution

      60/12.01=5.00 mol, which is 3.01×1024 atoms.

  34. NumericalHardNCERT Exercises 1.34

    A gas used to ripen fruit contains only carbon and hydrogen. Burning a sample completely gives 2.64 g of COX2 and 1.08 g of HX2O. Separately, 1.00 L of the gas at STP has a mass of 1.24 g. (Molar volume at STP 22.7 L mol−1; C 12.01, H 1.008, O 16.00)

    1. Find the empirical formula.

      Show solution

      C: 2.64/44.01=0.0600 mol. H: 2×1.08/18.02=0.120 mol. C : H =1:2, so the empirical formula is CHX2.

    2. Find the molar mass of the gas.

      g/mol
      Show solution

      M=1.24 g L−1×22.7 L mol−1=28.1 g mol−1.

    3. Find the molecular formula.

      Show solution

      n=28.1/14.03=2, so the gas is CX2HX4, ethene (ethylene).

  35. NumericalMediumNCERT Exercises 1.35

    Milk of magnesia neutralises stomach acid: Mg(OH)X2+2 HCl →MgClX2+2 HX2O. What mass of magnesium hydroxide (M=58.32 g mol−1) reacts completely with 30.0 mL of 0.10 M HCl? Give the answer in milligrams.

    mg
    Show solution

    n(HCl)=0.10×0.0300=3.0×10−3 mol. n(Mg(OH)X2)=1.5×10−3 mol. Mass =1.5×10−3×58.32=0.0875 g=87.5 mg.

  36. NumericalMediumNCERT Exercises 1.36

    Hydrogen is made in the lab by Zn+2 HCl→ZnClX2+HX2. What mass of HCl (36.46 g mol−1) reacts with 3.27 g of zinc (65.38 g mol−1)?

    g
    Show solution

    n(Zn)=3.27/65.38=0.0500 mol. n(HCl)=2×0.0500=0.100 mol. Mass =0.100×36.46=3.65 g.

  37. Quick answerMedium

    Assertion (A): The molality of a solution does not change when it is heated. Reason (R): Molality is defined using masses, and masses do not depend on temperature.

    Show answer

    Answer (A) Both A and R are true, and R correctly explains A

    Both are true, and R is exactly why A holds; molarity, which uses a volume, does change.
  38. Quick answerMedium

    What volume of oxygen at STP (molar volume 22.7 L mol−1) is needed to burn 1.6 g of methane completely? (CHX4+2 OX2→COX2+2 HX2O; M(CHX4)=16 g mol−1)

    L
    Show answer

    Answer 4.54 L

    n(CHX4)=0.10 mol, so n(OX2)=0.20 mol and V=0.20×22.7=4.5 L.
  39. Quick answerHard

    Assertion (A): 1.0 g of OX2 and 1.0 g of OX3 contain the same number of oxygen atoms. Reason (R): One gram of OX2 and one gram of OX3 contain the same number of molecules.

    Show answer

    Answer (C) A is true but R is false

    Each sample is 1.0 g of oxygen atoms, so both hold 1.0/16.00 mol of O atoms: A is true. But OX2 gives 1/32 mol and OX3 1/48 mol of molecules, so R is false.

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Quick check: MCQs on Some Basic Concepts of Chemistry

  1. Development of Chemistry2 questions
  2. 1.1 Importance of Chemistry1 question
  3. 1.2.2 Classification of Matter3 questions
  4. 1.3.7 Temperature3 questions
  5. 1.4.3 Dimensional Analysis3 questions
  6. 1.5.5 Avogadro's Law3 questions
  7. 1.6 Dalton's Atomic Theory2 questions
  8. 1.7.4 Formula Mass3 questions
  9. 1.8 Mole Concept and Molar Masses3 questions
  10. 1.9.1 Empirical Formula and Molecular Formula3 questions
  11. 1.10.1 Limiting Reagent3 questions
  12. 1.10.2 Reactions in Solutions3 questions
Glossary · 15

Key terms (English–Hindi)

accuracyयथार्थताyathaarthata
How close a result is to the true value.First seen in the chapter →
atomic mass unitपरमाणु द्रव्यमान मात्रकparmaanu dravyamaan maatrak
A twelfth of a carbon-12 atom's mass, 1.6605×10−24 g; symbol u.
compoundयौगिकyaugik
A pure substance of two or more elements combined chemically in a fixed ratio by mass.First seen in the chapter →
elementतत्वtatva
A pure substance made of only one kind of atom.First seen in the chapter →
empirical formulaमूलानुपाती सूत्रmoolaanupaati sootra
The simplest whole-number ratio of the atoms in a compound.First seen in the chapter →
limiting reagentसीमांत अभिकर्मकseemaant abhikarmak
The reactant used up first, which limits the amount of product.First seen in the chapter →
mixtureमिश्रणmishran
Two or more pure substances mixed in any ratio; separable by physical methods.First seen in the chapter →
molalityमोललताmolalta
Moles of solute per kilogram of solvent (m); it does not change with temperature.First seen in the chapter →
molar massमोलर द्रव्यमानmolar dravyamaan
How many grams one mole of a substance weighs; unit g mol−1.First seen in the chapter →
molarityमोलरताmolarta
Moles of solute per litre of solution (M); it changes with temperature.First seen in the chapter →
moleमोलmol
The SI unit of amount of substance: exactly 6.02214076×1023 entities.First seen in the chapter →
mole fractionमोल अंशmol ansh
The moles of one component divided by the total moles in the mixture.First seen in the chapter →
precisionपरिशुद्धताparishuddhata
How closely repeated measurements agree with each other.First seen in the chapter →
significant figuresसार्थक अंकsaarthak ank
The certain digits of a measurement plus one estimated digit.First seen in the chapter →
stoichiometryरससमीकरणमितिrassameekaranmiti
Calculating amounts of reactants and products from a balanced equation.First seen in the chapter →

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