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: 

∴ 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: 0.375 M aqueous solution means that 1000 mL of the solution contain sodium acetate = 0.375 mole

∴ 500 mL of the solution should contain sodium acetate = $\frac{0.375}{2}\backslash frac\{0.375\}\{2\}$ mole

Molar mass of sodium acetate = 82.0245 g mol⁻¹

∴ Mass of sodium acetate required
= $\frac{0.375}{2}\backslash frac\{0.375\}\{2\}$ mole × 82.0245 g mol⁻¹
= 15.380 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: Mass percent of 69% means that 100 g of nitric acid solution contain 69 g of nitric acid by mass.

Molar mass of nitric acid HNO₃ = 1 + 14 + 48 = 63 g mol⁻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: Empirical formula mass of Fe₂0₃ = 2 x 55.85 + 3 x 16.00 = 159.7 g mol^-1

Hence, molecular formula is same as empirical formula, viz.,Fe₂0₃.

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 

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 of methanol (CH₃OH) = 32 g mol⁻¹ = 0.032 kg mol⁻¹

Molarity of the given solution =

Applying M₁ × V₁ = M₂V₂
(Given solution)     (Solution to be prepared)

24.78 × V₁ = 0.25 × 2.5 L or V₁ = 0.02522 L = 25.22 mL

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: Pressure is the force (i.e., weight) acting per unit area But weight = mg

∴ Pressure = Weight per unit area =

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). 

Q1.15: Match the following prefixes with their multiples

Ans:

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

Ans: The digits in a properly recorded measurement are known as significant figures. It is also defined as follows. The total numbers of figures in a number including the last digit whose value is uncertain is called number of significant figures.

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) 15 ppm means 15 parts in million (10⁶) parts

∴ % by mass =  × 100 = 15 × 10⁻⁴ = 1.5 × 10⁻³ %

(ii) Molar mass of chloroform (CHCl₃) = 12 + 1 + 3 × 35.5 = 119.5 g mol⁻¹
100 g of the sample contain chloroform = 1.5 × 10⁻³ g

∴ 1000 g (1 kg) of the sample will contain chloroform = 1.5 × 10⁻² g

∴ Molality = 1.266 × 10⁻⁴ m

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

(iii) 1 mL = _______ = L = ______ dm³

Ans: (a)  Fixing the mass of dinitrogen as 28 g, masses of dioxygen combined will be 32,64, 32 and 80 g in the given four oxides. These'are in the ratio 1 : 2 : 1 : 5 which is a simple whole number ratio. Hence, the given data obey the law of multiple proportions.

(b)

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: 

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: H₂ and 0₂ react according to the equation

2H₂(g) + 0₂ (g) -->2H₂O (g) 

Thus, 2 volumes of H₂ react with 1 volume of 0₂ to produce 2 volumes of water vapour. Hence, 10 volumes of H₂ will react completely with 5 volumes of 0₂ to 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: 1 pm = 10^-12 m 
(i) 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: 

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) The least precise term has 3 significant figures (i.e., in 0.112). Hence, the answer should have 3 significant figures.
(ii) Leaving the exact number (5), the second term has 4 significant figures. Hence, the answer should have 4 significant figures.
(iii) In the given addition, the least number of decimal places in the term is 4. Hence, the answer should have 4 significant

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: 

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: 

Calculation of Empirical Formula

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: Step 1. To calculate mass of HCl in 25 mL of 0.75 m HCl
1000 mL of 0.75 M HCl contain HCl = 0.75 mol = 0.75 × 36.5 g = 24.375 g

∴ 25 mL of 0.75 HCl will contain HCl = 24.375/1000 × 25 g = 0.6844 g.

Step 2. To calculate mass of CaCO₃ reacting completely with 0.9125 g of HCl
CaCO₃ (s) + 2HCl (aq) -----> CaCl₂(aq) + CO₂(g) + H₂O

2 mol of HCl, i.e., 2 × 36.5 g = 73 g HCl react completely with CaCO₃ = 1 mol = 100 g

∴ 0.6844 g HCl will react completely with CaCO₃ = 100/73 × 0.6844 g = 0.938 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: 1 mole of MnO₂, i.e., 55 + 32 = 87 g MnO₂ react with 4 moles of HCl, i.e., 4 × 36.5 g = 146 g of HCl.