NCERT Solutions Class 11 Chemistry (Part-1) Chapter 1: Some Basic Concepts of Chemistry (Free PDF)

Class 11 Chemistry (Part-1) Chapter 1: Some Basic Concepts of Chemistry have introduced you with the importance and development of chemistry, its branches, the nature and scope of the subject, the states and classification of matter, as well as key concepts such as mass, volume, density, atomic and molecular masses, mole concept, and stoichiometry. This blog will provide you with exercises along with their solutions, which will help you understand the concepts more simply.

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NCERT Solutions Class 11 Chemistry (Part-1) Chapter 1: Some Basic Concepts of Chemistry

Below, we have provided you with exercises mentioned in the NCERT Class 11 Chemistry (Part-1) Chapter 1: Some Basic Concepts of Chemistry, along with their solutions.

Exercises

1. Calculate the molar mass of the following:

(i) H₂O (ii) CO₂ (iii) CH₄

2. Calculate the mass per cent of different elements present in sodium sulphate (Na₂SO₄)
3. Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.
4. Calculate the amount of carbon dioxide that could be produced when 1.5

(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.

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.
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 is 69%.
7. How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?
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.
9. Calculate the atomic mass (average) of chlorine using the following data:

	% Natural Abundance	Molar Mass
35Cl	75.77	34.9689
37Cl	24.23	36.9659

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.

11. What is the concentration of sugar (C₁₂H₂₂O₁₁ ) in mol L^–1 if 20 g are dissolved in

enough water to make a final volume of up to 2 L?

12. If the density of methanol is 0.793 kg L^–1, what volume of methanol is needed to make 2.5L of its 0.25 M solution?
13. Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal, is shown as: 1Pa = 1N m^–2

If the mass of air at sea level is 1034 g cm^–2, calculate the pressure in pascal.

14. What is the SI unit of mass? How is it defined?
15. What do you mean by significant figures?
16. A sample of drinking water was found to be severely contaminated with chloroform,

CHCl₃, supposed to be carcinogenic. 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.

17. Express the following in scientific notation:

(i) 0.0048

(ii) 234,000

(iii) 8008

(iv) 500.0

(v) 6.0012

18. 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

19. Round up the following up to three significant figures:

(i) 34.216

(ii) 10.4107

(iii) 0.04597

(iv) 2808

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

Mass of dinitrogen	Mass of dioxygen
14 g	16 g
14 g	32 g
28 g	32 g
28 g	80 g

Which law of chemical combination is obeyed by the above experimental data? Give

Its statement.

21. If the speed of light is 3.0 × 10 8 m s ^–1, calculate the distance covered by light in 2.00 ns.
22. 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

23. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:

N₂ (g) + H₂ (g)  2NH₃ (g)

(i) Calculate the mass of ammonia produced if 2.00 × 10 3 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?

24. How are 0.50 mol Na 2CO₃ and 0.50 M Na 2CO₃ different?
25. If 10 volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour would be produced?
26. Convert the following into basic units:

(i) 28.7 pm

(ii) 15.15 pm

(iii) 25365 mg

27. 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)

28. 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).
29. What will be the mass of one ¹²C atom in g?
30. How many significant figures should be present in the answer to the following calculations?

(i) 0.02856 × 298.15 × 0.112/0.5785

(ii) 5 × 5.364

(iii) 0.0125 + 0.7864 + 0.0215

31. 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.
32. 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.

33. Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the

reaction, CaCO₃ (s) + 2HCl (aq) → CaCl₂ (aq) + CO₂ (g) + H₂O (l)

What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?

34. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction

4HCl (aq) + MnO₂ (s) → 2H₂O (l) + MnCl₂ (aq) + Cl₂ (g)

How many grams of HCl react with 5.0 g of manganese dioxide?

Also Read: NCERT Solutions Class 11 Political Science Indian Constitution at Work Chapter 5: Legislature (Free PDF)

Solutions

1. Calculate the molar mass of:

(i) H₂O
= 2 × 1.008 + 16.00 = 18.016 g mol⁻¹

(ii) CO₂
= 12.01 + 2 × 16.00 = 44.01 g mol⁻¹

(iii) CH₄
= 12.01 + 4 × 1.008 = 16.042 g mol⁻¹

2. Mass percentage of different elements in Na₂SO₄

Molar mass of Na₂SO₄ = 2 × 22.99 + 32.06 + 4 × 16.00 = 142.04 g mol⁻¹

• Na: (2 × 22.99 / 142.04) × 100 = 32.39%
• S: (32.06 / 142.04) × 100 = 22.57%
• O: (64.00 / 142.04) × 100 = 45.04%

3. Empirical formula of iron oxide (69.9% Fe, 30.1% O)

• Moles of Fe = 69.9 / 55.85 ≈ 1.251
• Moles of O = 30.1 / 16.00 ≈ 1.881

Divide by the smallest:

• Fe = 1.251 / 1.251 = 1
• O = 1.881 / 1.251 ≈ 1.50
  Multiply both by 2 → Fe₂O₃

Empirical formula = Fe₂O₃

4.  CO₂ produced by burning carbon:

(i) 1 mol C + O₂ → CO₂ → 1 mol CO₂ = 44 g

(ii) 16 g O₂ = 0.5 mol O₂ → limits the reaction

 C + O₂ → CO₂ → 0.5 mol CO₂ = 0.5 × 44 = 22 g

(iii) 16 g O₂ (0.5 mol), 2 mol C → limits reaction again → CO₂ = 0.5 mol = 22 g

5. Sodium acetate for 500 mL of 0.375 M solution

Moles = Molarity × Volume (L) = 0.375 × 0.500 = 0.1875 mol
Mass = 0.1875 × 82.0245 = 15.38 g

6. Given:

• Density of solution = 1.41 g/mL = 1410 g/L
• Mass % of HNO₃ = 69%
• Molar mass of HNO₃ = 1 (H) + 14 (N) + 48 (3×16 O) = 63 g/mol

Step 1: Find mass of HNO₃ in 1 litre of solution
Mass of solution = 1000 mL × 1.41 g/mL = 1410 g
Mass of HNO₃ = 69% of 1410 =
= (69/100) × 1410 = 972.9 g

Step 2: Calculate the number of moles of HNO₃
Moles = Mass / Molar mass = 972.9 g / 63 g/mol = 15.44 mol

Answer: The concentration of nitric acid is 15.44 mol/L.

7. Given:

• Mass of CuSO₄ = 100 g
• Molar mass of CuSO₄ =
  Cu = 63.5, S = 32, O₄ = 4×16 = 64 → Total = 159.5 g/mol

Step 1: Find mass % of Cu in CuSO₄
= (63.5 / 159.5) × 100 = 39.8%

Step 2: Find the mass of copper in 100 g of CuSO₄
= (39.8 / 100) × 100 = 39.8 g

Answer: 39.8 g of copper can be obtained from 100 g of CuSO₄.

8. Given:

• Fe = 69.9%, O = 30.1%
• Atomic mass: Fe = 55.85, O = 16

Step 1: Calculate the moles of each element in 100 g
Fe: 69.9 / 55.85 = 1.251 mol
O: 30.1 / 16 = 1.881 mol

Step 2: Find the simplest mole ratio
Divide by the smallest:
Fe: 1.251 / 1.251 = 1
O: 1.881 / 1.251 ≈ 1.50

Multiply both by 2 to get whole numbers:
Fe : O = 2 : 3

Answer: The Empirical formula is Fe₂O₃ (which is also the molecular formula).

9. Step 1: Apply the formula for average atomic mass
   = (75.77/100) × 34.9689 + (24.23/100) × 36.9659
   = 0.7577 × 34.9689 + 0.2423 × 36.9659
   = 26.509 + 8.957 = 35.466 u

Answer: The average atomic mass of chlorine is 35.466 u.

10. Step 1: Calculate the molar mass of sugar (C₁₂H₂₂O₁₁)
    = (12×12.01) + (22×1.008) + (11×16.00)
    = 144.12 + 22.176 + 176.00 = 342.296 g/mol

Step 2: Moles of sugar =
= 20 g / 342.296 g/mol ≈ 0.05845 mol

Step 3: Molarity (mol/L) = moles / volume (L)
= 0.05845 mol / 2 L = 0.0292 mol/L

Answer: 0.0292 mol/L

11. Step 1: Moles of CH₃OH needed = Molarity × Volume =
    0.25 mol/L × 2.5 L = 0.625 mol

Step 2: Molar mass of methanol (CH₃OH):
= 12.01 + (4×1.008) + 16.00 = 32.042 g/mol

Mass required = 0.625 mol × 32.042 g/mol =
= 20.03 g = 0.02003 kg

Step 3: Volume = mass / density =
= 0.02003 kg / 0.793 kg/L = 0.02525 L = 25.25 mL

Answer: 25.25 mL

12. If the mass of air at sea level is 1034 g cm⁻², calculate the pressure in pascal.

Step 1: Convert mass per area to pressure
Force = mass × g = 1034 g/cm² × 9.8 m/s² =
= 1.034 kg/cm² × 9.8 = 10.1332 N/cm²

Now convert cm² to m²:
1 cm² = 1×10⁻⁴ m² → 10.1332 N/cm² =
= 10.1332 / (1×10⁻⁴) = 101332 Pa

Answer: Pressure = 1.013 × 10⁵ Pa

13. The SI unit of mass is the kilogram (kg). It is defined by taking the fixed numerical value of the Planck constant (h) as 6.62607015 × 10⁻³⁴ Js, which is equal to kg·m²·s⁻¹, where the metre and second are defined in terms of the speed of light and the caesium frequency.
14. Significant figures are all the meaningful digits in a measured or calculated quantity, including all digits known with certainty and one estimated digit. They indicate the precision of a measurement.
15. (i) 1 ppm = 1 part per million = 1 mg/kg = 1 mg in .1,000,000 mg
    So, 15 ppm = (15/1,000,000) × 100 = 0.0015%

Answer (i): 0.0015% by mass

(ii) Assume 1 kg of water (1000 g):

16. Mass of CHCl₃ = 15 mg = 0.015 g
    Molar mass of CHCl₃ = 12.01 + 1.008 + (3×35.45) = 119.37 g/mol

Molality = moles of solute/mass of solvent (kg)
Moles of CHCl₃ = 0.015 / 119.37 ≈ 1.257 × 10⁻⁴ mol
Mass of solvent = 1 kg

Answer (ii): Molality = 1.257 × 10⁻⁴ mol/kg

17. (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
18. (i) 0.0025 → 2
    (ii) 208 → 3
    (iii) 5005 → 4
    (iv) 126,000 → 3 (unless decimal is indicated)
    (v) 500.0 → 4
    (vi) 2.0034 → 5
19. (i) 34.216 → 34.2
    (ii) 10.4107 → 10.4
    (iii) 0.04597 → 0.0460
    (iv) 2808 → 2.81 × 10³ (or 2810 if not in scientific notation)
20. Let’s examine the ratios of masses of oxygen that combine with a fixed mass of dinitrogen.

Mass of N₂ is fixed: 14 g
Ratios of oxygen masses:

• Case 1: 16 g
• Case 2: 32 g → 32/16 = 2
• Case 3: N₂:O₂ = 28:32 → O₂ per 14 g N₂ = 16 g
  
• Case 4: 28:80 → O₂ per 14 g N₂ = 40 g

So: 16 g, 32 g, 40 g → Ratio: 16:32:40 = 2:4:5

21. Given:
    Speed of light, c = 3.0 × 10⁸ meters per second (m/s)
    Time, t = 2.00 nanoseconds (ns) = 2.00 × 10⁻⁹ seconds (s)

Calculation:
Distance = Speed × Time
= (3.0 × 10⁸ m/s) × (2.00 × 10⁻⁹ s)
= 6.00 × 10⁻¹ meters
= 0.60 meters

22. (i) 300 atoms of A + 200 molecules of B
    Reaction: A + B₂ → AB₂ → needs 1 A per 1 B₂ → Limiting: A (less B₂ needed)

(ii) 2 mol A + 3 mol B → 1 mol A + 1 mol B₂ → 2 mol A needs 2 mol B₂ → Only 1.5 mol B₂ available
So B is limiting
Limiting reagent: B

(iii) 100 atoms A + 100 molecules B → 100 atoms A needs 100 molecules B₂ →Limiting: B

(iv) 5 mol A + 2.5 mol B → 2.5 mol B₂ needed → 1 mol A per mol B₂

Limiting: Neither – exact stoichiometric ratio

No limiting reagent

(v) 2.5 mol A + 5 mol B → 5 mol B = 2.5 mol B₂ → 1:1 with A → Both exactly match

No limiting reagent

23. Balanced chemical equation:
    N₂ (g) + 3H₂ (g) → 2NH₃ (g)

Step 1: Calculate moles of each reactant

• Molar mass of N₂ = 28 g/mol
  Moles of N₂ = 2000 g ÷ 28 g/mol = 71.43 mol
• Molar mass of H₂ = 2 g/mol
  Moles of H₂ = 1000 g ÷ 2 g/mol = 500 mol

Step 2: Use the mole ratio from the balanced equation
According to the equation:
1 mol N₂ reacts with 3 mol H₂ to produce 2 mol NH₃

To react with 71.43 mol of N₂, we need:
71.43 × 3 = 214.29 mol of H₂

But we have 500 mol of H₂ → excess H₂
So N₂ is the limiting reagent

Step 3: Calculate moles of ammonia produced
1 mol N₂ produces 2 mol NH₃ → 71.43 mol N₂ will produce: 71.43 × 2 = 142.86 mol NH₃

Step 4: Calculate the mass of ammonia produced

Molar mass of NH₃ = 17 g/mol
Mass = 142.86 mol × 17 g/mol = 2428.62 g
Mass of ammonia produced = 2428.62 g or 2.43 × 10³ g

(ii) Will any of the two reactants remain unreacted?

Yes, hydrogen will remain unreacted since it is in excess.

(iii) Which one and how much of it remains unreacted?

Step 1: Hydrogen required = 214.29 mol

Step 2: Hydrogen available = 500 mol

Excess hydrogen = 500 − 214.29 = 285.71 mol

Mass of leftover hydrogen = 285.71 mol × 2 g/mol = 571.42 g

24. 0.50 mol Na₂CO₃ refers to the amount of sodium carbonate, i.e., 0.50 moles. It tells you the number of entities (molecules/ions), not the volume or concentration. 0.50 M Na₂CO₃ refers to the concentration of a solution, i.e., 0.50 moles of sodium carbonate dissolved in 1 litre of solution.
25. Balanced equation:
    2H₂ + O₂ → 2H₂O

This shows that:

2 volumes of hydrogen react with 1 volume of oxygen to give 2 volumes of water vapour.

Now, using the same ratio:

10 volumes of H₂ will react with 5 volumes of O₂ → will give 10 volumes of water vapour.

26. (i) 28.7 pm (picometres) 1 pm = 1 × 10⁻¹² m → 28.7 pm = 2.87 × 10⁻¹¹ m

(ii) 15.15 pm = 1.515 × 10⁻¹¹ m

(iii) 25365 mg (milligrams)
1 mg = 1 × 10⁻³ g = 1 × 10⁻⁶ kg → 25365 mg = 25.365 g = 0.025365 kg

27. Let’s calculate moles of each:

(i) 1 g of Au (gold)
Molar mass = 197 g/mol
Moles = 1 ÷ 197 = 0.00508 mol

(ii) 1 g of Na (sodium)
Molar mass = 23 g/mol
Moles = 1 ÷ 23 = 0.0435 mol

(iii) 1 g of Li (lithium)
Molar mass = 7 g/mol
Moles = 1 ÷ 7 = 0.143 mol

(iv) 1 g of Cl₂ (chlorine gas)
Molar mass = 70.9 g/mol
Moles = 1 ÷ 70.9 = 0.0141 mol
Each Cl₂ molecule has 2 atoms → Atoms = 0.0141 × 2 = 0.0282 mol atoms

Conclusion:
Lithium (Li) has the largest number of atoms because it has the highest number of moles per gram.

28. Assume total moles = 1
    Then:

• Moles of ethanol = 0.040
• Moles of water = 1 − 0.040 = 0.960

Mass of water = 0.960 mol × 18 g/mol = 17.28 g = 0.01728 kg
Assuming density of water = 1 g/mL, volume = 17.28 mL = 0.01728 L

Molarity = Moles of ethanol ÷ Volume of solution in litres
= 0.040 ÷ 0.01728 = 2.31 mol/L

Final Answer:
Molarity = 2.31 M

29. 1 mole of ¹²C atoms = 12 g
    Number of atoms in 1 mole = 6.022 × 10²³

Mass of 1 atom = 12 g ÷ 6.022 × 10²³
= 1.993 × 10⁻²³ g

30. (i) (0.02856 × 298.15 × 0.112) ÷ 0.5785

Multiply the numbers: → Result ≈ 0.165

Lowest significant figures in input = 3 (0.112)
Answer = 0.165 (3 significant figures)

(ii) 5 × 5.364 = 26.82

Since “5” has 1 significant figure, the final answer should have 1 significant figure

Answer = 30

(iii) 0.0125 + 0.7864 + 0.0215 = 0.8204

The number with the least decimal places is 0.0125 (4 decimals) → Keep 4 decimal places in the result
Answer = 0.8204

31. 52 moles of Ar
    Atoms = 52 × Avogadro’s number = 52 × 6.022 × 10²³ = 3.13 × 10²⁵ atoms

(ii) 52 u of He
1 atom of He = 4 u

52 u = 52 ÷ 4 = 13 atoms
 

(iii) 52 g of He
Molar mass = 4 g/mol
Moles = 52 ÷ 4 = 13 mol
Atoms = 13 × 6.022 × 10²³ = 7.83 × 10²⁴ atoms

32. Given:

• CO₂ formed = 3.38 g → C = (12/44) × 3.38 = 0.922 g
• H₂O formed = 0.690 g → H = (2/18) × 0.690 = 0.077 g

Moles:

• C = 0.922 ÷ 12 = 0.0768
• H = 0.077 ÷ 1 = 0.077

Ratio ≈ C:H = 1:1 → Empirical formula = CH

Given: 10 L of gas weighs 11.6 g at STP
Molar volume at STP = 22.4 L
Moles = 10 ÷ 22.4 = 0.446 mol
Molar mass = 11.6 ÷ 0.446 = ~26 g/mol

Empirical formula mass of CH = 13
Molecular formula = (26 ÷ 13) = 2 → Molecular formula = C₂H₂

33. Reaction:
    CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

Step 1: Moles of HCl
M = 0.75 mol/L, Volume = 25 mL = 0.025 L
Moles of HCl = 0.75 × 0.025 = 0.01875 mol

Step 2: Moles of CaCO₃ required
1 mol CaCO₃ reacts with 2 mol HCl → Moles of CaCO₃ = 0.01875 ÷ 2 = 0.009375 mol

Step 3: Mass = moles × molar mass
Molar mass of CaCO₃ = 100 g/mol
Mass = 0.009375 × 100 = 0.938 g

34. Reaction:
    MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O

Step 1: Moles of MnO₂
Molar mass = 86.9 g/mol
Moles = 5.0 ÷ 86.9 = 0.0575 mol

Step 2: Moles of HCl required
1 mol MnO₂ requires 4 mol HCl → HCl = 4 × 0.0575 = 0.23 mol

Step 3: Mass of HCl = moles × molar mass = 0.23 × 36.5 = 8.40 g

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