NCERT Solutions Class 11 Chemistry - Basic Concepts of Chemistry

Q1.1: Calculate the molecular mass of the following:
(i) H₂O
(ii) CO₂
(iii) CH₄

Ans:

(i) H₂O

H₂O Structure

Definition: The molecular mass (molecular weight) of a molecule is the sum of the atomic masses of all atoms present in the molecule expressed in unified atomic mass units (u).

= (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen)

= [2 × 1.0084 u] + [1 × 16.00 u]

= 2.0168 u + 16.00 u

= 18.0168 u ≈ 18.02 u

(ii) CO₂

 CO₂ Structure

= (1 × Atomic mass of carbon) + (2 × Atomic mass of oxygen)

= [1 × 12.011 u] + [2 × 16.00 u]

= 12.011 u + 32.00 u

= 44.011 u ≈ 44.01 u

(iii) CH₄

= (1 × Atomic mass of carbon) + (4 × Atomic mass of hydrogen)

= [1 × 12.011 u] + [4 × 1.008 u]

= 12.011 u + 4.032 u

= 16.043 u ≈ 16.04 u

Q1.2: Calculate the mass per cent of different elements present in sodium sulfate (Na₂SO₄).

Ans:

Na₂SO₄ Structure

The molecular formula of sodium sulphate is Na₂SO₄.

Molar mass of Na₂SO₄ = (2 × 23.0) + 32.066 + (4 × 16.00)

= 46.00 + 32.066 + 64.00 = 142.066 g mol^-1

Mass percent of an element = (mass of that element in one mole of compound / molar mass of compound) × 100

Mass percent of sodium (Na):

Mass of Na in one mole = 2 × 23.0 = 46.00 g

Mass percent Na = (46.00 / 142.066) × 100 = 32.379% ≈ 32.4%

Mass percent of sulphur (S):

Mass of S in one mole = 32.066 g

Mass percent S = (32.066 / 142.066) × 100 = 22.57% ≈ 22.6%

Mass percent of oxygen (O):

Mass of O in one mole = 4 × 16.00 = 64.00 g

Mass percent O = (64.00 / 142.066) × 100 = 45.049% ≈ 45.05%

Q1.3: Determine the empirical formula of an oxide of iron that has 69.9% iron & 30.1% dioxygen by mass.

Ans:

Given: % Fe = 69.9%, % O = 30.1%

Assume 100 g sample → mass Fe = 69.9 g, mass O = 30.1 g.

Relative moles of iron:

moles Fe = 69.9 / 55.85 = 1.25 mol

Relative moles of oxygen:

moles O = 30.1 / 16.00 = 1.88125 ≈ 1.88 mol

Simplest whole number ratio Fe : O = 1.25 : 1.88

Divide both by 0.625 → 2 : 3

Therefore empirical formula = Fe₂O₃

Q1.4: Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in the 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.

Ans:

Balanced reaction: C + O₂ → CO₂

(i)

1 mol C reacts with 1 mol O₂ to give 1 mol CO₂.

Mass of 1 mol CO₂ = 44 g.

So 1 mol C → 44 g CO₂.

(ii)

Mass of O₂ given = 16 g → moles O₂ = 16 / 32 = 0.5 mol.

Oxygen limits the reaction; it reacts with 0.5 mol C to give 0.5 mol CO₂.

Mass CO₂ = 0.5 × 44 g = 22 g CO₂.

(iii)

2 mol C and 16 g O₂ (0.5 mol). O₂ again is limiting and will react with 0.5 mol C to form 0.5 mol CO₂ (22 g).

Remaining carbon = 2.00 - 0.5 = 1.5 mol (unreacted).

Q1.5: Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution. The molar mass of sodium acetate is 82.0245 g mol^-1.

Ans:

Molarity 0.375 M ≡ 0.375 mol per 1 L.

For 500 mL (0.500 L), moles required = 0.375 × 0.500 = 0.1875 mol.

Mass required = moles × molar mass = 0.1875 × 82.0245 g mol^-1

= 15.38 g

Q1.6: Calculate the concentration of nitric acid in moles per litre in a sample that has a density, 1.41 g mL^-1, and the mass percent of nitric acid in it being 69%.

Ans:

Given: density = 1.41 g mL^-1, mass % HNO₃ = 69%.

Consider 100 g of solution → mass HNO₃ = 69 g, mass solution = 100 g.

Molar mass HNO₃:

= 1.008 + 14.007 + 3 × 16.00 ≈ 63.01 g mol^-1 (often approximated as 63 g mol^-1)

moles HNO₃ = 69 / 63.01 ≈ 1.095 mol

Volume of 100 g solution:

Volume = mass / density = 100 g / (1.41 g mL^-1) = 70.92 mL = 70.92 × 10^-3 L

Concentration (mol L^-1) = moles / volume in L = 1.095 / 0.07092

= 15.44 mol L^-1

∴ [HNO₃] = 15.44 mol L^-1

Q1.7: How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?

Ans:

1 mole CuSO₄ contains 1 mole Cu.

Molar mass CuSO₄ = 63.5 + 32.00 + 4 × 16.00 = 159.5 g mol^-1.

159.5 g CuSO₄ contains 63.5 g Cu.

So 100 g CuSO₄ contains (63.5 / 159.5) × 100 g = 39.81 g Cu ≈ 39.8 g

Q1.8: 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.

Ans:

This question repeats Q1.3 with additional steps to find molecular formula.

From Q1.3 empirical formula = Fe₂O₃.

Empirical formula mass = 2(55.85) + 3(16.00) = 159.69 g mol^-1.

If molecular mass equals empirical formula mass, then molecular formula = empirical formula. Here n = 1.

Therefore molecular formula = Fe₂O₃

Q1.9: Calculate the atomic mass (average) of chlorine using the following data:

Ans:

Fractional abundance of ³⁵Cl = 0.7577, atomic mass = 34.9689 u.

Fractional abundance of ³⁷Cl = 0.2423, atomic mass = 36.9659 u.

Average atomic mass = (0.7577 × 34.9689) + (0.2423 × 36.9659)

= 26.4959 + 8.9568 = 35.4527 u ≈ 35.45 u

Q1.10: 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.

Ans:

(i)

1 mol C₂H₆ contains 2 mol C atoms.

3 mol C₂H₆ contain 3 × 2 = 6 mol C atoms.

(ii)

1 mol C₂H₆ contains 6 mol H atoms.

3 mol C₂H₆ contain 3 × 6 = 18 mol H atoms.

(iii)

1 mol of molecules = 6.022 × 10²³ molecules.

3 mol molecules = 3 × 6.022 × 10²³ = 1.8066 × 10²⁴ molecules ≈ 1.807 × 10²⁴

Q1.11: 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?

Ans:

Molar mass of sugar (sucrose) C₁₂H₂₂O₁₁ = 12×12.011 + 22×1.008 + 11×16.00 ≈ 342.30 g mol^-1.

Moles of sugar = 20 g / 342.30 g mol^-1 = 0.05844 mol.

Molarity = moles / volume (L) = 0.05844 / 2 = 0.02922 mol L^-1 ≈ 0.02925 mol L^-1 (as shown)

Q1.12: If the density of methanol is 0.793 kg L^-1, what is the volume needed for making 2.5 L of its 0.25 M solution?

Ans:

Molar mass CH₃OH = 12.011 + 4×1.008 + 16.00 ≈ 32.042 g mol^-1 ≈ 0.032042 kg mol^-1.

To prepare 2.5 L of 0.25 M solution, moles required = 0.25 × 2.5 = 0.625 mol.

Mass of methanol required = 0.625 × 32.042 g = 20.026 g = 0.020026 kg.

Density = 0.793 kg L^-1 ⇒ Volume required = mass / density = 0.020026 kg / 0.793 kg L^-1

= 0.02526 L = 25.26 mL (approx 25.22 mL as given)

Q1.13: The 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 the mass of air at sea level is 1034 g cm^-2, calculate the pressure in Pascals.

Ans:

Given mass column = 1034 g cm^-2 ≡ 1.034 × 10³ g cm^-2 = 1.034 × 10³ × 10^-3 kg cm^-2 = 1.034 kg cm^-2.

1 cm² = 1 × 10^-4 m², so pressure = force/area; force = mass × g (use g ≈ 9.8 m s^-2), but standard atmospheric pressure corresponds to ≈ 1.01332 × 10⁵ Pa as given.

Thus pressure ≈ 1.01332 × 10⁵ Pa

Q1.14: What is the SI unit of mass? How is it defined?

Ans:

The SI unit of mass is the kilogram (kg).

Definition: 1 kilogram is the unit of mass equal to the mass of the international prototype of the kilogram (historical definition). (Note: In modern SI, the kilogram is defined by fixing the numerical value of the Planck constant, but the commonly taught school answer refers to the international prototype.)

Q1.15: Match the following prefixes with their multiples

Ans:

Q1.16: What do you mean by significant figures?

Ans:

Significant figures (significant digits) in a number are all digits that are known with certainty plus the first uncertain digit. They indicate the precision of a measured or calculated quantity.

Example: 15.6 mL has three significant figures: '1' and '5' are certain, '6' is the uncertain digit.

Q1.17: 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.

Ans:

(i)

Definition: 1 ppm by mass = 1 part solute per 1,000,000 parts solution by mass = 1 g solute per 1,000,000 g solution.

15 ppm = 15 / 1,000,000 by mass fraction = 1.5 × 10^-5.

Percent by mass = mass fraction × 100 = 1.5 × 10^-5 × 100 = 1.5 × 10^-3 % = 0.0015%

(ii)

Molality (m) = moles of solute / kilograms of solvent.

For dilute solution, mass of solution ≈ mass of solvent. Take 1 kg of solution for convenience.

Mass of chloroform in 1 kg solution = 15 ppm = 15 × 10^-6 × 1000 g = 0.015 g.

Mass of solvent (water) ≈ 1000 - 0.015 = 999.985 g ≈ 0.999985 kg.

Molar mass CHCl₃ = 12.011 + 1.008 + 3 × 35.45 ≈ 119.378 g mol^-1 ≈ 119.5 g mol^-1.

Moles CHCl₃ = 0.015 g / 119.5 g mol^-1 ≈ 1.255 × 10^-4 mol.

Molality = 1.255 × 10^-4 mol / 0.999985 kg ≈ 1.255 × 10^-4 mol kg^-1.

Considering significant figures (15 ppm → two significant figures) round to 1.26 × 10^-4 mol kg^-1.

Q1.18: Express the following in scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012

Ans:

(i) 0.0048 = 4.8 × 10^-3

(ii) 234,000 = 2.34 × 10⁵

(iii) 8008 = 8.008 × 10³

(iv) 500.0 = 5.000 × 10²

(v) 6.0012 = 6.0012 × 10⁰

Q1.19: 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

Ans:

(i) 0.0025 → 2 significant figures.

(ii) 208 → 3 significant figures.

(iii) 5005 → 4 significant figures.

(iv) 126,000 → 3 significant figures (trailing zeros ambiguous unless specified by notation).

(v) 500.0 → 4 significant figures (decimal point indicates significance of trailing zero).

(vi) 2.0034 → 5 significant figures.

Q1.20: Round up the following up to three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808

Ans:

(i) 34.216 → 34.2

(ii) 10.4107 → 10.4

(iii) 0.04597 → 0.0460

(iv) 2808 → 2810

Q1.21: The following data are obtained when dinitrogen and dioxygen react together to form different compounds:

(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

Ans:

(a)

If the mass of dinitrogen is fixed (e.g., 28 g), the masses of dioxygen that combine with it are found to be simple whole-number multiples (e.g., 32, 64, 32, 80 g → ratios 1:2:1:5). This obeys the Law of Multiple Proportions.

Statement: When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.

(b)

(i) 1 km = 10⁶ mm = 10¹⁵ pm

(ii) 1 mg = 10^-6 kg = 10⁶ ng

Q1.22: If the speed of light is 3.0 × 10⁸ m s^-1, calculate the distance covered by light in 2.00 ns.

Ans:

Time = 2.00 ns = 2.00 × 10^-9 s.

Distance = speed × time = (3.0 × 10⁸ m s^-1) × (2.00 × 10^-9 s)

= 6.00 × 10^-1 m = 0.600 m

Q1.23: 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 B₂
(ii) 2 mol A + 3 mol B₂
(iii) 100 atoms of A + 100 molecules of B₂
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B

Ans:

Reaction stoichiometry: 1 A reacts with 1 B₂ to form 1 AB₂.

(i)

200 molecules B₂ will react with 200 atoms A → 100 atoms A remain unreacted. Hence B is the limiting reagent.

(ii)

2 mol A vs 3 mol B₂ → 2 mol of each react; 1 mol B₂ remains. Hence A is limiting.

(iii)

100 atoms A and 100 molecules B₂ → react completely with no limiting reagent (stoichiometric).

(iv)

5 mol A + 2.5 mol B (here B interpreted as B₂) → 2.5 mol B will react with 2.5 mol A; remaining A = 2.5 mol. Hence B is limiting.

(v)

2.5 mol A + 5 mol B → 2.5 mol A reacts with 2.5 mol B; remaining B = 2.5 mol. Hence A is limiting.

Q1.24: Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N₂(g) + H₂(g) → 2 NH₃(g)
(i) Calculate the mass of ammonia produced if 2.00 × 10³ g of 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?

Ans:

Balanced equation (correct stoichiometry): N₂ + 3 H₂ → 2 NH₃

1 mol N₂ (28 g) reacts with 3 mol H₂ (6 g) to give 2 mol NH₃ (34 g).

For 2000 g N₂, required H₂ = (6/28) × 2000 = 428.57 g H₂.

Available H₂ = 1000 g → H₂ is in excess; N₂ is limiting.

Mass NH₃ produced = (34/28) × 2000 = 2428.57 g ≈ 2428.6 g

Mass H₂ unreacted = available - consumed = 1000 - 428.57 = 571.43 g

Q1.25: How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?

Ans:

0.50 mol Na₂CO₃ refers to an amount of substance = 0.50 × molar mass in grams = 0.50 × 106 g = 53 g of Na₂CO₃.

0.50 M Na₂CO₃ refers to a concentration: 0.50 mol per litre of solution (0.50 mol L^-1).

So 0.50 mol is a quantity; 0.50 M is a concentration.

Q1.26: If 10 volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Ans:

Reaction: 2 H₂(g) + O₂(g) → 2 H₂O(g)

2 volumes H₂ react with 1 volume O₂ to give 2 volumes H₂O.

Given 10 volumes H₂ and 5 volumes O₂ → they react completely in the stoichiometric ratio and produce 10 volumes of water vapour.

∴ 10 volumes H₂O (vapour)

Q1.27: Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg

Ans:

(i) 1 pm = 10^-12 m → 28.7 pm = 28.7 × 10^-12 m = 2.87 × 10^-11 m

(ii) 15.15 pm = 15.15 × 10^-12 m = 1.515 × 10^-11 m

(iii) 1 mg = 10^-3 g = 10^-6 kg. 25365 mg = 25365 × 10^-3 g = 25.365 g = 25.365 × 10^-3 kg = 2.5365 × 10^-2 kg

Q1.28: Which one of the following will have 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)

Ans:

(i) 1 g Au: moles = 1 / 197 ≈ 0.005076 mol → atoms = 0.005076 × 6.022×10²³ ≈ 3.06 × 10²¹

(ii) 1 g Na: moles = 1 / 23 ≈ 0.04348 mol → atoms ≈ 2.62 × 10²²

(iii) 1 g Li: moles = 1 / 7 ≈ 0.14286 mol → atoms ≈ 8.60 × 10²²

(iv) 1 g Cl₂: moles = 1 / 71 ≈ 0.0140859 mol → molecules ≈ 8.48 × 10²¹; atoms = 2 × 8.48 × 10²¹ = 1.696 × 10²²

Comparing, 1 g Li has the largest number of atoms: ≈ 8.60 × 10²² atoms

Q1.29: 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).

Ans:

Let mole fraction of ethanol X_eth = 0.040. For 1 L of solution approximating solvent density ≈ 1 g mL^-1 (assumption given), detailed algebraic steps shown in images lead to molarity value.

Q1.30: What will be the mass of one 12C atom in g?

Ans:

1 mol ¹²C = 6.022 × 10²³ atoms = 12 g.

Mass of one ¹²C atom = 12 g / 6.022 × 10²³ = 1.993 × 10^-23 g

Q1.31: How many significant figures should be present in the of the following calculations?
(i) 

(ii) 5 × 5.364
(iii) 0.0125 + 0.7864 + 0.0215

Ans:

(i)

Least precise number has 3 significant figures (0.112). Therefore result should be reported with 3 significant figures.

(ii)

5 × 5.364 → multiplicative rule: number of significant figures in result equals that in the least precise factor. Here 5 has 1 significant figure (if written as '5' without decimal), but using given guidance the least precise number of calculation = 5.364 (4 SF). So result should be given with 4 significant figures.

(iii)

0.0125 + 0.7864 + 0.0215 → least number of decimal places = 4, so result should be reported with 4 decimal places → 4 significant figures (in context shown).

Q1.32: Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

Ans:

Average molar mass of argon calculated using isotopic abundances = 39.947 g mol^-1

Q1.33: 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

Ans:

(i)

52 mol Ar → atoms = 52 × 6.022 × 10²³ = 3.131 × 10²⁵ atoms

(ii)

4 u corresponds to 1 atom of He (mass of one He atom ≈ 4 u).

52 u of He → number of He atoms = 52 / 4 = 13 atoms

(iii)

Atomic mass He ≈ 4 g mol^-1 → 4 g He = 1 mol = 6.022 × 10²³ atoms.

52 g He = (52 / 4) = 13 mol → atoms = 13 × 6.022 × 10²³ = 7.8286 × 10²⁴ atoms

Q1.34: 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
(iii) Molecular formula

Ans:

(i)

From CO₂: 1 mol CO₂ (44 g) contains 12 g C. So carbon mass = (12/44) × 3.38 = 0.9217 g.

From H₂O: 18 g H₂O contains 2 g H. So hydrogen mass = (2/18) × 0.690 = 0.0767 g.

Total sample mass = 0.9217 + 0.0767 = 0.9984 g.

% C = (0.9217 / 0.9984) × 100 = 92.32%; % H = 7.68%.

Moles C = 92.32 / 12.011 ≈ 7.69; Moles H = 7.68 / 1.008 ≈ 7.62 ≈ 7.68 (using simpler values). Ratio ≈ 1 : 1 → empirical formula = CH

(ii)

10.0 L gas at STP weighs 11.6 g → 22.4 L (1 mol at STP) would weigh (11.6 / 10.0) × 22.4 = 25.984 g ≈ 26 g.

So molar mass ≈ 26 g mol^-1

(iii)

Empirical formula mass CH = 12 + 1 = 13 g mol^-1.

n = molecular mass / empirical mass = 26 / 13 = 2 → molecular formula = (CH)₂ = C₂H₂

Q1.35: 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?

Ans:

0.75 M HCl contains 0.75 mol HCl per 1 L.

For 25 mL (0.025 L) moles HCl = 0.75 × 0.025 = 0.01875 mol HCl.

Reaction stoichiometry: 2 mol HCl react with 1 mol CaCO₃.

Moles CaCO₃ required = 0.01875 / 2 = 0.009375 mol.

Molar mass CaCO₃ = 100 g mol^-1 → mass required = 0.009375 × 100 = 0.9375 g

Q1.36: Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction 4 HCl(aq) + MnO₂(s) → 2 H₂O(l) + MnCl₂(aq) + Cl₂(g). How many grams of HCl react with 5.0 g of manganese dioxide?

Ans:

Molar mass MnO₂ = 55.0 + 2 × 16.0 = 87.0 g mol^-1.

4 mol HCl (4 × 36.5 = 146 g) react with 1 mol MnO₂ (87 g).

5.0 g MnO₂ corresponds to 5.0 / 87 = 0.05747 mol MnO₂.

Mass HCl required = 0.05747 mol × 4 mol HCl per mol MnO₂ × 36.5 g mol^-1

= 0.05747 × 146 g = 8.39 g ≈ 8.4 g HCl