Complete Textbook Solutions On Chapter 1 Some Basic Concepts of Chemistry

MAHARASHTRA BOARD · CLASS 11

Some Basic Concepts of Chemistry

Chapter 1 — Complete Textbook Solutions & In-Text Exercises

📘 MCQs ✍️ Short Answers 🔬 Examples 🧮 Numericals 📖 Explain 📝 Textbook Exercises

📘 1. Choose the Most Correct Option

A

A sample of pure water, whatever the source, always contains ___ by mass of oxygen and 11.1% by mass of hydrogen.

• a. 88.9
• b. 18
• c. 80
• d. 16

Show Answer

a. 88.9

B

Which of the following compounds can NOT demonstrate the law of multiple proportions?

• a. NO, NO₂
• b. CO, CO₂
• c. H₂O, H₂O₂
• d. Na₂S, NaF

Show Answer

d. Na₂S, NaF

C

Which temperature reads the same value on the Celsius and Fahrenheit scales?

• a. – 40°
• b. + 40°
• c. – 80°
• d. – 20°

Show Answer

a. – 40°

D

SI unit of the quantity electric current is

• a. Volt
• b. Ampere
• c. Candela
• d. Newton

Show Answer

b. Ampere

E

In N₂ + 3H₂ → 2NH₃, the ratio by volume of N₂, H₂ and NH₃ is 1:3:2. This illustrates the law of

• a. definite proportion
• b. reciprocal proportion
• c. multiple proportion
• d. gaseous volumes

Show Answer

d. gaseous volumes

F

Which of the following has the maximum number of molecules?

• a. 7 g N₂
• b. 2 g H₂
• c. 8 g O₂
• d. 20 g NO₂

Show Answer

b. 2 g H₂

G

How many grams of H₂O are present in 0.25 mol of it?

• a. 4.5
• b. 18
• c. 0.25
• d. 5.4

Show Answer

a. 4.5

H

The number of molecules in 22.4 cm³ of nitrogen gas at STP is

• a. 6.022 × 10²⁰
• b. 6.022 × 10²³
• c. 22.4 × 10²⁰
• d. 22.4 × 10²³

Show Answer

a. 6.022 × 10²⁰

I

Which of the following has the largest number of atoms?

• a. 1 g Au(s)
• b. 1 g Na(s)
• c. 1 g Li(s)
• d. 1 g Cl₂(g)

Show Answer

c. 1 g Li(s)

✍️ 2. Answer the Following Questions

A

State and explain Avogadro's law.

Show Answer

In 1811, Avogadro proposed that "Equal volumes of all gases at the same temperature and pressure contain equal number of molecules."

Hydrogen(g) + Oxygen(g) -> Water(g) [2 vol] [1 vol] [2 vol] [2n molecules] [n molecules] [2n molecules]

i.e., 2 molecules of hydrogen combine with 1 molecule of oxygen to give 2 molecules of water vapour.

B

Point out the difference between 12 g of carbon and 12 u of carbon.

Show Answer

12 g of carbon is the molar mass of carbon; 12 u of carbon is the mass of one carbon atom.

C

How many grams does an atom of hydrogen weigh?

Show Answer

1.6736 × 10^-24 g.

D

Calculate the molecular mass (in u) of NH₃, CH₃COOH, and C₂H₅OH.

Show Answer

NH₃: (1×14.0) + (3×1.0) = 17 u

CH₃COOH: (2×12.0) + (4×1.0) + (2×16.0) = 60 u

C₂H₅OH: (2×12.0) + (6×1.0) + (1×16.0) = 46 u

E

How many particles are present in 1 mole of a substance?

Show Answer

6.0221367 × 10²³ particles.

F

What is the SI unit of amount of a substance?

Show Answer

Mole (mol).

G

What is meant by molar volume of a gas?

Show Answer

The volume occupied by one mole of a gas at STP (0 °C, 1 atm), which is 22.4 dm³.

H

State and explain the law of conservation of mass.

Show Answer

• Mass can neither be created nor destroyed during a chemical reaction.
• Lavoisier's combustion experiments (phosphorus, mercury) showed the weight gained by the solid equalled the weight lost by the air.
• Total mass of reactants = total mass of products.

I

State the law of multiple proportions.

Show Answer

When two elements A and B form more than one compound, the masses of B that combine with a fixed mass of A are always in a ratio of small whole numbers.

🔬 3. Give One Example of Each

Term	Example
Homogeneous mixture	Solution — e.g., aqueous sugar solution
Heterogeneous mixture	Suspension — e.g., sand in water
Element	Gold
Compound	Distilled water

🧮 4. Solve Problems

A

Ratio of molecules in 1 mole of NH₃ and 1 mole of HNO₃.

Show Solution

Both contain 6.022 × 10²³ molecules per mole.

Ans: Ratio = 1 : 1

B

Moles of hydrogen in 0.448 L of hydrogen gas at STP.

Show Solution

n = Volume at STP ÷ 22.4 L mol^-1 = 0.448 ÷ 22.4

Ans: 0.02 mol

C

Mass of an atom of hydrogen is 1.008 u. Find the mass of 18 atoms.

Show Solution

18 × 1.008 u

Ans: 18.144 u

D

Number of atoms in (a) 254 u and (b) 254 g of iodine (I = 127 u).

Show Solution

a. 254 u ÷ 127 u = 2 atoms

b. Moles = 254/127 = 2 mol → atoms = 2 × 6.022×10²³ = 1.2044 × 10²⁴ atoms

E

A carbon pencil writing weighs 5 mg. Find (a) moles of carbon, (b) atoms of carbon in 12 mg.

Show Solution

a. 5×10^-3 g ÷ 12 g mol^-1 = 4.167 × 10^-4 mol

b. 12×10^-3 g ÷ 12 = 1×10^-3 mol → atoms = 6.022 × 10²⁰

F

250 g of glucose (C₆H₁₂O₆) costs Rs 40. Find the cost per mole.

Show Solution

Molar mass of glucose = 180 g mol^-1

Cost/mole = (40 × 180) ÷ 250

Ans: Rs 28.8 per mole

G

Natural abundance: ¹⁰B = 19.60% (mass 10.13), ¹¹B = 80.40% (mass 11.009). Find average atomic mass of boron.

Show Solution

[(10.13×19.60) + (11.009×80.40)] ÷ 100

Ans: 10.84 u

H

Convert to Fahrenheit: (a) 40 °C, (b) 30 °C.

Show Solution

°F = (9/5)(°C) + 32

a. (9/5)(40)+32 = 104 °F   b. (9/5)(30)+32 = 86 °F

I

Moles and molecules of acetic acid in 22 g of it.

Show Solution

Molar mass CH₃COOH = 60 g mol^-1; moles = 22/60 = 0.367 mol

Molecules = 0.367 × 6.022×10²³ = 2.210 × 10²³

J

24 g carbon + oxygen → 88 g CO₂. Find mass of oxygen used.

Show Solution

Carbon + Oxygen -> Carbon dioxide 12 g 32 g 44 g

Scaling ×2: 24 g C + 64 g O₂ → 88 g CO₂

Ans: 64 g oxygen used

K

Number of atoms in (a) 0.4 mol N, (b) 1.6 g S. (N=14u, S=32u)

Show Solution

a. 0.4 × 6.022×10²³ = 2.4088 × 10²³ atoms

b. 1.6/32 = 0.05 mol → 0.05 × 6.022×10²³ = 3.011 × 10²² atoms

L

2.0 g metal + O₂ → 3.2 g oxide. 1.42 g metal + steam → 2.27 g oxide. Which law is verified?

Show Solution

Reaction 1: % O = 1.2/3.2 × 100 = 37.5%. Reaction 2: % O = 0.85/2.27 × 100 ≈ 37.5%.

Same proportion regardless of source → Law of definite proportions is verified.

M

In 2 moles of acetaldehyde (CH₃CHO), find moles of C, H, O and number of molecules.

Show Solution

C: 2×2=4 mol   H: 2×4=8 mol   O: 2×1=2 mol

Molecules: 2×6.022×10²³ = 12.044 × 10²³

N

Moles of MgO in (i) 80 g, (ii) 10 g. (Mg=24, O=16)

Show Solution

Molar mass = 40 g mol^-1. i. 80/40 = 2 mol   ii. 10/40 = 0.25 mol

O

Volume of CO₂ at STP for (i) 5 mol, (ii) 0.5 mol.

Show Solution

i. 5×22.4 = 112 dm³   ii. 0.5×22.4 = 11.2 dm³

P

Mass of KClO₃ needed to liberate 6.72 dm³ O₂ at STP. (Molar mass KClO₃=122.5)

Show Solution

2KClO3 -> 2KCl + 3O2 [2 mol=245g] [3 mol=67.2 dm3]

x = (245 × 6.72) ÷ 67.2

Ans: 24.5 g

Q

Atoms of H, N, C, O in 5.6 g of urea, (NH₂)₂CO.

Show Solution

Molar mass = 60 g mol^-1; moles = 5.6/60 = 0.0933 mol (4H, 2N, 1C, 1O per molecule)

H = 2.247×10²³   N = 1.124×10²³   C = 0.562×10²³   O = 0.562×10²³

R

Mass of SO₂ from burning 16 g sulfur in excess oxygen. (S=32u, O=16u)

Show Solution

Sulphur + Oxygen -> Sulphur dioxide 32 g 32 g 64 g

Scaling ×0.5: 16 g S + 16 g O₂ → 32 g SO₂

Ans: 32 g

📖 5. Explain

A

The need for the term "average atomic mass".

Show Answer

Many elements exist as a mixture of isotopes with different atomic masses. The average atomic mass — weighted by isotopic abundance — is used to represent such elements accurately.

B

Molar mass.

Show Answer

The mass of one mole of a substance in grams. It's numerically equal to the atomic/molecular/formula mass in u (e.g., H₂O = 18.0 u → 18.0 g mol^-1).

C

Mole concept.

Show Answer

A mole is the amount of substance containing as many entities as atoms in exactly 12 g of carbon-12 — that is, 6.0221367 × 10²³ particles.

D

Formula mass, with an example.

Show Answer

Sum of atomic masses in a formula unit — used for ionic lattices like NaCl, which have no discrete molecules.

Example: NaCl = 23.0 u + 35.5 u = 58.5 u

E

Molar volume of a gas.

Show Answer

One mole of any gas occupies 22.4 dm³ at STP (Avogadro's law). n = Volume at STP ÷ 22.4 dm³ mol^-1.

IUPAC has since redefined STP pressure to 1 bar; under this newer standard, molar volume = 22.71 L mol^-1.

F

Types of matter (on the basis of chemical composition).

Show Answer

Pure substances — fixed composition. Split into Elements (metals, nonmetals, metalloids) and Compounds.

Mixtures — no fixed composition, separable physically. Split into Homogeneous (e.g., solutions) and Heterogeneous (e.g., suspensions).

📝 In-Text Textbook Exercises

1

Classify as mixture or pure substance: sea water, gasoline, skin, a rusty nail, a textbook page, diamond.

Show Answer

All are mixtures except diamond, which is a pure substance.

2

Classify as element or compound: mercuric oxide, helium, water, table salt, iodine, mercury, oxygen, nitrogen.

Show Answer

Compounds: mercuric oxide, water, table salt.

Elements: helium, iodine, mercury, oxygen, nitrogen.

3

10 volumes of H₂ react with 5 volumes of O₂ — how many volumes of water vapour form?

Show Answer

2H2(g) + O2(g) -> 2H2O(g) [2 vol] [1 vol] [2 vol]

Ans: 10 volumes of water vapour

4

What is an atom and a molecule? Order of magnitude of an atom's mass? What are isotopes?

Show Answer

• An atom is the smallest indivisible particle of an element.
• A molecule is two or more atoms held together by chemical bonds.
• Mass of one atom is of the order 10^-27 kg.
• Isotopes are atoms of the same element with the same atomic number but different mass numbers.

5

Find the formula mass of CaSO₄ (Ca=40.1u, S=32.1u, O=16.0u).

Show Answer

40.1 + 32.1 + (4×16.0) = 136.2 u

6

One dozen and one gross mean how many items?

Show Answer

Dozen = 12; Gross = 144.

7

Volume in dm³ occupied by 60.0 g of ethane at STP.

Show Answer

Molar mass ethane = 30 g mol^-1 → moles = 2 mol → Volume = 2×22.4

Ans: 44.8 dm³

📚 Maharashtra Board · Class 11 Chemistry · Chapter 1 — Some Basic Concepts of Chemistry