Chemistry Basics

NCERT Solution 3

Question 19: How many significant figures are present in the following?

  1. 0.0025
  2. 208
  3. 5005
  4. 126,000
  5. 500.0
  6. 2.0034

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


Question 20: Round up the following upto three significant figures:

  1. 34.216
  2. 10.4107
  3. 0.04597
  4. 2808

Answer: (i) 34.2, (ii) 10.4, (iii) 0.0460 (iv) 2810

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

Mass of dinitrogenMass of dioxygen
14 g16 g
14 g32 g
28 g32 g
28 g80 g

(a) Which law of chemical combination is obeyed by the above experimental data? Give its statement.

Answer: Law of multiple proportions

(b) Fill in the blanks in the following conversions:

(i) 1 km = ...................... mm = ...................... pm

Answer: 1 km = 106 mm = 1015 pm

(ii) 1 mg = ...................... kg = ...................... ng

Answer: 1 mg = 10-6 kg = 106 ng

(iii) 1 mL = ...................... L = ...................... dm3

Answer: 1 mL = 10-3 L = 10-3 dm3


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

Answer: 1 ns = 10-9 s

So, distance in 1 ns

= 3.0 × 108 m × 1-1 × 2 × 10-9

= 6.0 × 10-1 m = 0.6 m

Question 23: In a reaction

A + B2 → AB2

Identify the limiting reagent, if any, in the following reaction mixtures.

  1. 300 atoms of A + 200 molecules of B
  2. 2 mol A + 3 mol B
  3. 100 atoms of A + 100 molecules of B
  4. 5 mol A + 2.5 mol B
  5. 2.5 mol A + 5 mol B

Answer: As per equation, 1 atom of A reacts with 1 molecule of B

So, in case (i) 200 molecules of B will react with 200 atoms of A. This will leave 100 atoms of A. This means, A is the limiting reagent in this situation. Similarly, limiting reagents in other options are as follows:

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

Question 24: Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:

N2 (g) + 3H2 (g) → 2NH3 (g)

  1. Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 × 103 g of dihydrogen.
  2. Will any of the two reactants remain unreacted?
  3. If yes, which one and what would be its mass?

Answer: 1 mole of nitrogen (28 g) reacts with 3 mole of hydrogen (6 g)

So, 2000 g nitrogen will react with `6/(28)×2000=428.6` g hydrogen

It will produce 2428.6 g ammonia

This will leave 1000 – 428.6 = 571.4 g hydrogen unreacted


Question 25: How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?

Answer: Molar mass of sodium carbonate

= 2 × 23 + 12 + 3 × 16 = 46 + 12 + 48 = 106 g

0.5 mol of sodium carbonate = 106 ÷ 2 = 53 g

0.5 M means 1 liter solution contains 0.53 g of sodium carbonate.

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

Answer: Following equation shows reaction between hydrogen and oxygen to produce water

2H2 + O2 → 2H2

It means that 2 volume of hydrogen reacts with 1 volume of oxygen to produce 2 volume of water

So, 10 volume of water vapour would be produced

Question 27: Convert the following into basic units:

(i) 28.7 pm

Answer: 28.7 pm = 28.7 × 10-12 m = 2.87 × 10-11 m

(ii) 15.15 μs

Answer: 15.15 μs = 15.15 × 10-6 s = 1.515 × 10-5 s

(iii) 25365 mg

Answer: 25365 mg = 25365 × 10-6 kg = 2.5365 × 10-2 kg



Copyright © excellup 2014