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NCERT Solutions · Class 11 Chemistry · NCERT Chemistry Part I and II Part I · Chapter 1

Chapter 1: Some Basic Concepts of Chemistry (Basic Concepts)

Step-by-step solutions to all 36 exercises of Unit 1, Some Basic Concepts of Chemistry (NCERT Class 11 Chemistry Part I, 2026-27): molar mass and mass per cent, empirical and molecular formulae, mole concept, molarity, molality and mole fraction, limiting reagent, significant figures, laws of chemical combination and stoichiometry. All 36 questions are answered, with the key answer highlighted.

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Moles $n = \frac{m}{M} = \frac{N}{N _{A}}$ , $N_{A} = 6.022 \times 1 0^{23}$ mol⁻¹. Molarity = mol of solute per L of solution; molality = mol per kg of solvent; mole fraction $x_{A} = \frac{n _{A}}{n _{A} + n _{B}}$ . Empirical formula from the simplest mole ratio; molecular formula $= n \times$ empirical, $n = \frac{M}{EF mass}$ . Find the limiting reagent before calculating a product. Atomic masses: H 1.008, C 12.01, N 14.01, O 16.00, Na 22.99, S 32.06, Cl 35.45, Fe 55.85, Cu 63.55.

Exercises

1.1

Calculate the molar mass of the following: (i) H₂O (ii) CO₂ (iii) CH₄

Solution

(i) $2 (1.008) + 16.00 = 18.02$ (ii) $12.01 + 2 (16.00) = 44.01$ (iii) $12.01 + 4 (1.008) = 16.04$

(i) 18.02 g mol⁻¹ (ii) 44.01 g mol⁻¹ (iii) 16.04 g mol⁻¹

1.2

Calculate the mass per cent of different elements present in sodium sulphate (Na₂SO₄).

Solution

Molar mass $= 2 (22.99) + 32.06 + 4 (16.00) = 142.04$ g mol⁻¹.

Na: $\frac{45.98}{142.04} \times 100 = 32.37%$ ; S: $\frac{32.06}{142.04} \times 100 = 22.57%$ ; O: $\frac{64.00}{142.04} \times 100 = 45.06%$ .

Na 32.37%, S 22.57%, O 45.06%

1.3

Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

Solution

In 100 g: Fe $= \frac{69.9}{55.85} = 1.25$ mol; O $= \frac{30.1}{16.00} = 1.88$ mol.

Divide by the smaller: Fe : O = 1 : 1.5 = 2 : 3.

Fe₂O₃

1.4

Calculate the amount of carbon dioxide that could be produced when (i) 1 mole of carbon is burnt in air. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. (iii) 2 moles of carbon are burnt in 16 g of dioxygen.

Solution

$C + O X_{2} CO X_{2}$ : 1 mol C + 1 mol O₂ → 1 mol CO₂ (44 g).

(i) Air has excess O₂, so 1 mol C gives 1 mol CO₂ = 44 g.

(ii) 16 g O₂ = 0.5 mol: O₂ is the limiting reagent; 0.5 mol CO₂ = 22 g.

(iii) Again only 0.5 mol O₂: 0.5 mol CO₂ = 22 g (carbon is in excess).

(i) 44 g (ii) 22 g (iii) 22 g

1.5

Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol⁻¹.

Solution

$n = M \times V = 0.375 \times 0.500 = 0.1875$ mol; mass $= 0.1875 \times 82.0245$ .

15.38 g

1.6

Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL⁻¹ and the mass per cent of nitric acid in it being 69%.

Solution

1 L of solution has mass 1410 g, containing $0.69 \times 1410 = 972.9$ g HNO₃ (M = 63.02 g mol⁻¹).

Moles in 1 L $= \frac{972.9}{63.02}$

15.44 mol L⁻¹

1.7

How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?

Solution

M(CuSO₄) $= 63.55 + 32.06 + 64.00 = 159.61$ g mol⁻¹; each mole contains 63.55 g Cu.

Cu from 100 g $= \frac{63.55}{159.61} \times 100$

39.8 g

1.8

Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively.

Solution

Empirical formula Fe₂O₃ (Exercise 1.3); empirical formula mass $= 2 (55.85) + 3 (16.00) = 159.7$ g mol⁻¹. The molar mass of this oxide (iron(III) oxide) is also 159.7 g mol⁻¹, so $n = 1$ .

Fe₂O₃

1.9

Calculate the atomic mass (average) of chlorine using the following data: ³⁵Cl: natural abundance 75.77%, molar mass 34.9689; ³⁷Cl: 24.23%, 36.9659.

Solution

$0.7577 \times 34.9689 + 0.2423 \times 36.9659 = 26.4959 + 8.9568$

35.45 u

1.10

In three moles of ethane (C₂H₆), calculate the following: (i) Number of moles of carbon atoms. (ii) Number of moles of hydrogen atoms. (iii) Number of molecules of ethane.

Solution

(i) $3 \times 2 = 6$ mol (ii) $3 \times 6 = 18$ mol (iii) $3 \times 6.022 \times 1 0^{23} = 1.807 \times 1 0^{24}$

(i) 6 mol (ii) 18 mol (iii) 1.807 × 10²⁴ molecules

1.11

What is the concentration of sugar (C₁₂H₂₂O₁₁) in mol L⁻¹ if its 20 g are dissolved in enough water to make a final volume up to 2 L?

Solution

M $= 12 (12.01) + 22 (1.008) + 11 (16.00) = 342.3$ g mol⁻¹; $n = \frac{20}{342.3} = 0.0584$ mol.

Concentration $= \frac{0.0584}{2}$

0.0292 mol L⁻¹

1.12

If the density of methanol is 0.793 kg L⁻¹, what is its volume needed for making 2.5 L of its 0.25 M solution?

Solution

$n = 0.25 \times 2.5 = 0.625$ mol; mass $= 0.625 \times 32.04 = 20.03$ g $= 0.02003$ kg.

Volume $= \frac{0.02003 kg}{0.793 kg L ^{- 1}} = 0.0253$ L.

25.3 mL

1.13

Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below: 1 Pa = 1 N m⁻². If mass of air at sea level is 1034 g cm⁻², calculate the pressure in pascal.

Solution

On 1 cm² the air has mass 1.034 kg: force $= 1.034 \times 9.8 = 10.13$ N on $1 0^{- 4}$ m².

$P = \frac{10.13}{1 0 ^{- 4}}$

1.01 × 10⁵ Pa

1.14

What is the SI unit of mass? How is it defined?

Solution

The kilogram (kg). It is defined by fixing the numerical value of the Planck constant h to be $6.62607015 \times 1 0^{- 34}$ when expressed in J s (kg m² s⁻¹), with the metre and second defined in terms of c and the caesium frequency. (Earlier it was the mass of the international prototype kilogram.)

Kilogram, defined through the fixed value of the Planck constant.

1.15

Match the following prefixes with their multiples: (i) micro (ii) deca (iii) mega (iv) giga (v) femto — 10⁶, 10⁹, 10⁻⁶, 10⁻¹⁵, 10

Solution

micro → 10⁻⁶; deca → 10; mega → 10⁶; giga → 10⁹; femto → 10⁻¹⁵

1.16

What do you mean by significant figures?

Solution

Significant figures are the meaningful digits in a measured quantity: all the digits known with certainty plus one uncertain (estimated) last digit. They show the precision of the measurement. For example, 11.2 mL has three significant figures.

Certain digits + one uncertain digit of a measurement.

1.17

A sample of drinking water was found to be severely contaminated with chloroform, CHCl₃, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass). (i) Express this in per cent by mass. (ii) Determine the molality of chloroform in the water sample.

Solution

(i) 15 ppm = 15 g in $1 0^{6}$ g of solution: $\frac{15}{1 0 ^{6}} \times 100 = 1.5 \times 1 0^{- 3} %$ .

(ii) M(CHCl₃) $= 12.01 + 1.008 + 3 (35.45) = 119.4$ g mol⁻¹; 15 g $= 0.1257$ mol in (practically) $1 0^{6}$ g $= 1 0^{3}$ kg of water.

Molality $= \frac{0.1257}{1 0 ^{3}} = 1.26 \times 1 0^{- 4}$ mol kg⁻¹.

(i) 1.5 × 10⁻³ % (ii) 1.26 × 10⁻⁴ m

1.18

Express the following in the scientific notation: (i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012

Solution

(i) 4.8 × 10⁻³ (ii) 2.34 × 10⁵ (iii) 8.008 × 10³ (iv) 5.000 × 10² (v) 6.0012 × 10⁰

1.19

How many significant figures are present in the following? (i) 0.0025 (ii) 208 (iii) 5005 (iv) 126,000 (v) 500.0 (vi) 2.0034

Solution

(i) Leading zeros are not significant. (ii) and (iii) Zeros between non-zero digits count. (iv) Trailing zeros without a decimal point do not count. (v) Trailing zero after a decimal point counts. (vi) All digits count.

(i) 2 (ii) 3 (iii) 4 (iv) 3 (v) 4 (vi) 5

1.20

Round up the following upto three significant figures: (i) 34.216 (ii) 10.4107 (iii) 0.04597 (iv) 2808

Solution

(i) 34.2 (ii) 10.4 (iii) 0.0460 (iv) 2810

1.21

The following data are obtained when dinitrogen and dioxygen react together to form different compounds: (i) 14 g N₂ with 16 g O₂ (ii) 14 g with 32 g (iii) 28 g with 32 g (iv) 28 g with 80 g. (a) Which law of chemical combination is obeyed by the above experimental data? Give its statement. (b) Fill in the blanks in the following conversions: (i) 1 km = ... mm = ... pm (ii) 1 mg = ... kg = ... ng (iii) 1 mL = ... L = ... dm³

Solution

(a) For a fixed 28 g of N₂, the masses of O₂ are 32, 64, 32 and 80 g, in the ratio 2 : 4 : 2 : 5, a simple whole-number ratio. This is the law of multiple proportions: if two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers.

(b) (i) $1 km = 1 0^{6}$ mm $= 1 0^{15}$ pm (ii) $1 mg = 1 0^{- 6}$ kg $= 1 0^{6}$ ng (iii) $1 mL = 1 0^{- 3}$ L $= 1 0^{- 3}$ dm³

(a) Law of multiple proportions (b) 10⁶ mm, 10¹⁵ pm; 10⁻⁶ kg, 10⁶ ng; 10⁻³ L, 10⁻³ dm³

1.22

If the speed of light is 3.0 × 10⁸ m s⁻¹, calculate the distance covered by light in 2.00 ns.

Solution

$d = 3.0 \times 1 0^{8} \times 2.00 \times 1 0^{- 9}$

0.600 m

1.23

In a reaction A + B₂ → AB₂, identify the limiting reagent, if any, in the following reaction mixtures. (i) 300 atoms of A + 200 molecules of B (ii) 2 mol A + 3 mol B (iii) 100 atoms of A + 100 molecules of B (iv) 5 mol A + 2.5 mol B (v) 2.5 mol A + 5 mol B

Solution

A and B₂ react in a 1 : 1 ratio; the one present in the smaller amount is limiting.

(i) 200 B₂ for 300 A: B limiting. (ii) 2 mol A for 3 mol B₂: A limiting. (iii) Exactly 1 : 1: no limiting reagent. (iv) 2.5 mol B₂ for 5 mol A: B limiting. (v) 2.5 mol A for 5 mol B₂: A limiting.

(i) B (ii) A (iii) none (iv) B (v) A

1.24

Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: N₂ (g) + 3H₂ (g) → 2NH₃ (g). (i) Calculate the mass of ammonia produced if 2.00 × 10³ g dinitrogen reacts with 1.00 × 10³ g of dihydrogen. (ii) Will any of the two reactants remain unreacted? (iii) If yes, which one and what would be its mass?

Solution

$n (N X_{2}) = \frac{2000}{28} = 71.4$ mol; $n (H X_{2}) = \frac{1000}{2} = 500$ mol. 71.4 mol N₂ needs $3 \times 71.4 = 214.3$ mol H₂, which is available: N₂ is the limiting reagent.

(i) NH₃ $= 2 \times 71.4 = 142.9$ mol $\times 17 = 2.43 \times 1 0^{3}$ g.

(ii) Yes. (iii) Dihydrogen: $500 - 214.3 = 285.7$ mol $= 571$ g left.

(i) 2.43 × 10³ g (ii) Yes (iii) H₂, about 5.7 × 10² g

1.25

How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?

Solution

0.50 mol is an amount of substance: $0.50 \times 106 = 53$ g of Na₂CO₃. 0.50 M is a concentration: 0.50 mol (53 g) of Na₂CO₃ dissolved in 1 L of solution.

0.50 mol = 53 g of Na₂CO₃; 0.50 M = 53 g per litre of solution.

1.26

If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Solution

$2 H X_{2} + O X_{2} 2 H X_{2} O$ : by Gay Lussac's law, volumes are in the ratio 2 : 1 : 2. So 10 vol H₂ + 5 vol O₂ give 10 vol of water vapour.

10 volumes

1.27

Convert the following into basic units: (i) 28.7 pm (ii) 15.15 pm (iii) 25365 mg

Solution

(i) 2.87 × 10⁻¹¹ m (ii) 1.515 × 10⁻¹¹ m (iii) 2.5365 × 10⁻² kg

1.28

Which one of the following will have the largest number of atoms? (i) 1 g Au (s) (ii) 1 g Na (s) (iii) 1 g Li (s) (iv) 1 g of Cl₂(g)

Solution

Moles of atoms in 1 g: Au $\frac{1}{197} = 0.0051$ ; Na $\frac{1}{22.99} = 0.0435$ ; Li $\frac{1}{6.94} = 0.144$ ; Cl (as atoms) $\frac{1}{35.45} = 0.0282$ . The largest is Li (the lightest atom).

(iii) 1 g Li

1.29

Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

Solution

Take 1 L (1000 g) of water $= \frac{1000}{18} = 55.5$ mol. $x = \frac{n}{n + 55.5} = 0.040 \Rightarrow n = \frac{0.040 \times 55.5}{0.960} = 2.31$ mol.

Taking the solution volume as about 1 L (dilute solution), molarity ≈ 2.31 M.

About 2.31 M

1.30

What will be the mass of one ¹²C atom in g?

Solution

$\frac{12 g}{6.022 \times 1 0 ^{23}}$

1.99 × 10⁻²³ g

1.31

How many significant figures should be present in the answer of the following calculations? (i) (0.02856 × 298.15 × 0.112)/0.5785 (ii) 5 × 5.364 (iii) 0.0125 + 0.7864 + 0.0215

Solution

(i) Multiplication/division: as many as the least precise factor, 0.112 (3). (ii) 5 is an exact number, so 5.364 decides: 4. (iii) Addition: as many decimal places as the least, all have 4: sum 0.8204 (4 significant figures).

(i) 3 (ii) 4 (iii) 4

1.32

Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes: ³⁶Ar 35.96755 g mol⁻¹, 0.337%; ³⁸Ar 37.96272 g mol⁻¹, 0.063%; ⁴⁰Ar 39.9624 g mol⁻¹, 99.600%.

Solution

$0.00337 \times 35.96755 + 0.00063 \times 37.96272 + 0.99600 \times 39.9624 = 0.1212 + 0.0239 + 39.8026$

39.948 g mol⁻¹

1.33

Calculate the number of atoms in each of the following (i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He.

Solution

(i) $52 \times 6.022 \times 1 0^{23} = 3.131 \times 1 0^{25}$ (ii) one He atom is 4 u: $\frac{52}{4} = 13$ atoms (iii) $\frac{52}{4} = 13$ mol $= 13 \times 6.022 \times 1 0^{23} = 7.83 \times 1 0^{24}$

(i) 3.131 × 10²⁵ (ii) 13 (iii) 7.83 × 10²⁴

1.34

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.

Solution

C in 3.38 g CO₂ $= 3.38 \times \frac{12}{44} = 0.922$ g $= 0.0768$ mol. H in 0.690 g H₂O $= 0.690 \times \frac{2}{18} = 0.0767$ g $= 0.0767$ mol.

(i) C : H = 1 : 1, so CH.

(ii) 10.0 L weigh 11.6 g, so 22.4 L weigh $\frac{11.6 \times 22.4}{10.0} = 26.0$ g: M = 26 g mol⁻¹.

(iii) EF mass = 13; $n = \frac{26}{13} = 2$ : C₂H₂ (ethyne).

(i) CH (ii) 26 g mol⁻¹ (iii) C₂H₂

1.35

Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction, CaCO₃ (s) + 2 HCl (aq) → CaCl₂ (aq) + CO₂(g) + H₂O(l). What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?

Solution

$n (HCl) = 0.025 \times 0.75 = 0.01875$ mol; CaCO₃ needed $= \frac{1}{2} \times 0.01875 = 0.009375$ mol.

Mass $= 0.009375 \times 100.09$

0.94 g

1.36

Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction 4 HCl (aq) + MnO₂(s) → 2H₂O (l) + MnCl₂(aq) + Cl₂ (g). How many grams of HCl react with 5.0 g of manganese dioxide?

Solution

$n (MnO X_{2}) = \frac{5.0}{86.94} = 0.0575$ mol; HCl $= 4 \times 0.0575 = 0.230$ mol $\times 36.46$ .

8.4 g

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