Chapter 1 Some Basic Concepts of Chemistry Solutions
Question - 21 : - The following data are obtained when dinitrogen anddioxygen react together to form different compounds:
Mass of dinitrogen Mass ofdioxygen
(i) 14g 16 g
(ii) 14g 32 g
(iii) 28 g 32 g
(iv) 28g 80 g
(a) Which law of chemical combination is obeyed by theabove 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
Answer - 21 : -
(a)
If we fix the mass of dinitrogen at 28 g, then the massesof dioxygen that will combine with the fixed mass of dinitrogen are 32 g, 64 g,32 g, and 80 g.
The masses of dioxygen bear a whole number ratio of1:2:2:5. Hence, the given experimental data obeys the law of multipleproportions. The law states that if two elements combine to form more than onecompound, then the masses of one element that combines with the fixed mass ofanother element are in the ratio of small whole numbers.
(b)
(i) 1 km = 1 km ×
1 km = 106 mm
1 km = 1 km ×
1 km = 1015 pm Hence, 1 km = 106 mm = 1015 pm
(ii) 1 mg = 1 mg ×
⇒ 1 mg = 10–6 kg
1 mg = 1 mg × 
⇒ 1 mg = 106 ng
1 mg = 10–6 kg= 106 ng
(iii) 1 mL = 1 mL × 
⇒ 1 mL = 10–3 L
1 mL = 1 cm3 = 1 cm3
⇒ 1 mL = 10–3 dm3
1 mL = 10–3 L= 10–3 dm3
If the speed of light is 3.0 ×108 ms-1, calculate the distance covered by light in 2.00 ns.
Answer - 22 : -
According to the question:
Time taken to cover the distance = 2.00 ns
= 2.00 × 10–9 s
Speed of light = 3.0 × 108 ms–1
Distance travelled by light in 2.00 ns
= Speed of light × Time taken
= (3.0 × 108 ms –1)(2.00 × 10–9 s)
= 6.00 × 10–1 m
= 0.600 m
Question - 23 : - In a reaction
A + B2 → AB2
Identify the limiting reagent, if any, in the followingreaction 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
Answer - 23 : -
A limiting reagent determines the extent of a reaction. Itis the reactant which is the first to get consumed during a reaction, therebycausing the reaction to stop and limiting the amount of products formed.
(i) According to the given reaction, 1 atom of A reacts with1 molecule of B. Thus, 200 molecules of B will react with 200 atoms of A,thereby leaving 100 atoms of A unused. Hence, B is the limiting reagent.
(ii) According to the reaction, 1 mol of A reacts with 1mol of B. Thus, 2 mol of A will react with only 2 mol of B. As a result, 1 molof B will not be consumed. Hence, A is the limiting reagent.
(iii) According to the given reaction, 1 atom of Acombines with 1 molecule of B. Thus, all 100 atoms of A will combine with all100 molecules of B. Hence, the mixture is stoichiometric where no limitingreagent is present.
(iv) 1 mol of atom A combines with 1 mol of molecule B.Thus, 2.5 mol of B will combine with only 2.5 mol of A. As a result, 2.5 mol ofA will be left as such. Hence, B is the limiting reagent.
(v) According to the reaction, 1 mol of atom A combineswith 1 mol of molecule B. Thus, 2.5 mol of A will combine with only 2.5 mol ofB and the remaining 2.5 mol of B will be left as such. Hence, A is the limitingreagent.
Question - 24 : - Dinitrogen and dihydrogen react with each other to produceammonia according to the following chemical equation:
N2(g) + H2(g) →2NH3(g)
(i) Calculate the mass of ammonia producedif 2.00 × 103 g dinitrogen reacts with 1.00 × 103 gof dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
Answer - 24 : -
(i) Balancing the given chemical equation,
From the equation, 1 mole (28 g) of dinitrogen reacts with3 mole (6 g) of dihydrogen to give 2 mole (34 g) of ammonia.
⇒2.00 × 103 g of dinitrogen will react with 
dihydrogen i.e.,
2.00 × 103 g of dinitrogenwill react with 428.6 g of dihydrogen.
Given,
Amount of dihydrogen = 1.00 × 103 g
Hence, N2 is the limitingreagent.
28 g of N2 produces34 g of NH3.
Hence, mass of ammonia produced by 2000 g of N2
= 2428.57 g
(ii) N2 is the limitingreagent and H2 is the excess reagent. Hence, H2 willremain unreacted.
(iii) Mass of dihydrogen left unreacted =1.00 × 103 g – 428.6 g
= 571.4 g
Question - 25 : - How are 0.50 mol Na2CO3 and0.50 M Na2CO3 different?
Answer - 25 : -
Molar mass of Na2CO3 =(2 × 23) + 12.00 + (3× 16)
= 106 g mol–1
Now, 1 mole of Na2CO3 means106 g of Na2CO3.
0.5mol of Na2CO3 
= 53 g Na2CO3
⇒ 0.50 M of Na2CO3 =0.50 mol/L Na2CO3
Hence, 0.50 mol of Na2CO3 ispresent in 1 L of water or 53 g of Na2CO3 is presentin 1 L of water.
If ten volumes of dihydrogen gas react with five volumesof dioxygen gas, how many volumes of water vapour would be produced?
Answer - 26 : -
Reaction of dihydrogen with dioxygen can be written as:
Now, two volumes of dihydrogen react with one volume ofdioxygen to produce two volumes of water vapour.
Hence, ten volumes of dihydrogen will react with fivevolumes of dioxygen to produce ten volumes of water vapour.
Question - 27 : - Which one of the following will have largest number ofatoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of Cl2(g)
Answer - 27 : -
1 g of Au (s) 
mol of Au (s)
atoms of Au (s)
= 3.06 × 1021atoms of Au (s)
1 g of Na (s) =
mol of Na (s)
atoms of Na (s)= 0.262 × 1023 atoms of Na(s)
= 26.2 × 1021 atoms of Na(s)
1 g of Li (s) 
mol of Li (s)
atoms of Li (s)= 0.86 × 1023 atoms of Li(s)
= 86.0 × 1021 atoms of Li(s)
1 g of Cl2 (g) 
mol of Cl2 (g)(Molar mass of Cl2 molecule= 35.5 × 2 = 71 g mol–1)
molecules of Cl2 (g)
= 0.0848 × 1023 moleculesof Cl2 (g)
= 8.48 × 1021 molecules ofCl2 (g)
As one molecule of Cl2 containstwo atoms of Cl.
Number of atoms of Cl = 2× 8.48 × 1021 =16.96× 1021 atoms of Cl
Hence, 1 g of Li (s) will have the largest number ofatoms.
Question - 28 : - Calculate the molarity of a solution of ethanol in waterin which the mole fraction of ethanol is 0.040 (assume the density of water tobe one).
Answer - 28 : - Mole fraction of C2H5OH 
Number of moles present in 1 L water:
Substituting the value of 
in equation (1),
Molarity of solution 
= 2.314 M
What will be the mass of one 12Catom in g?
Answer - 29 : -
1 mole of carbon atoms = 6.023 × 1023 atomsof carbon
= 12 g of carbon
Massof one 12C atom 
Question - 30 : - How many significant figures should be present in theanswer of the following calculations?
(i) 
(ii) 5 × 5.364
(iii) 0.0125 + 0.7864 + 0.0215
Answer - 30 : -
(i)
Least precise number of calculation = 0.112
Number of significant figures in the answer
= Number of significant figures in the least precisenumber
= 3
(ii) 5 × 5.364
Least precise number of calculation = 5.364
Number of significant figures in the answer = Number ofsignificant figures in 5.364
= 4
(iii) 0.0125 + 0.7864 + 0.0215
Since the least number of decimal places in each term isfour, the number of significant figures in the answer is also 4.