Chapter 1 Some Basic Concepts of Chemistry Solutions
Question - 31 : - Use the data given in the following table tocalculate the molar mass of naturally occurring argon isotopes:
| Isotope | Isotopic molar mass | Abundance |
| 36Ar | 35.96755 gmol–1 | 0.337% |
| 38Ar | 37.96272 gmol–1 | 0.063% |
| 40Ar | 39.9624 gmol–1 | 99.600% |
Answer - 31 : -
Molar mass of argon
= 39.947 gmol–1
Question - 32 : - 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.
Answer - 32 : -
(i) 1 mole of Ar = 6.022 × 1023 atoms of Ar
52 mol of Ar = 52 × 6.022 × 1023 atomsof Ar
= 3.131 × 1025 atoms of Ar
(ii) 1 atom of He = 4 u of He
Or,
4 u of He = 1 atom of He
1 u of He
atom of He52u of He 
atom of He= 13 atoms of He
(iii) 4 g of He = 6.022 × 1023 atoms of He
.52 g of He
atoms of He
= 7.8286 × 1024 atoms of He
Question - 33 : - 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 ofwater and no other products. A volume of 10.0 L (measured at STP) of thiswelding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii)molar mass of the gas, and (iii) molecular formula.
Answer - 33 : -
(i) 1 mole (44 g) of CO2 contains 12 g ofcarbon.
3.38 g of CO2 willcontain carbon
= 0.9217 g
18 g of water contains 2 g of hydrogen.
0.690 g of water willcontain hydrogen 
= 0.0767 g
Since carbon and hydrogen are the only constituents of thecompound, the total mass of the compound is:
= 0.9217 g + 0.0767 g
= 0.9984 g
Percent of C in thecompound 
= 92.32%
Percent of H in the compound 
= 7.68%
Moles of carbon in the compound 
= 7.69Moles of hydrogen in the compound = 
= 7.68
Ratio of carbon to hydrogen in the compound = 7.69: 7.68
= 1: 1
Hence, the empirical formula of the gas is CH.
(ii) Given,
Weight of 10.0L of the gas (at S.T.P) = 11.6 g
Weight of 22.4 L of gas atSTP 
= 25.984 g
≈ 26 g
Hence, the molar mass of the gas is 26 g.
(iii) Empirical formula mass of CH = 12 + 1 = 13 g
n = 2
Molecular formula of gas = (CH)n
= C2H2
Calcium carbonate reacts with aqueous HCl togive CaCl2 and CO2 according to the reaction, CaCO3(s) +2 HCl(aq) → CaCl2(aq) +CO2(g) + H2O(l)
What mass of CaCO3 isrequired to react completely with 25 mL of 0.75 M HCl?
Answer - 34 : -
0.75 M of HCl ≡ 0.75 mol of HCl are present in 1 L ofwater
≡ [(0.75 mol) × (36.5 g mol–1)]HCl is present in 1 L of water
≡ 27.375 g of HCl is present in 1 L of water
Thus, 1000 mL of solution contains 27.375 g of HCl.
Amount of HCl present in 25 mL of solution
= 0.6844 g
From the given chemical equation,
2 mol of HCl (2 × 36.5 = 73 g) react with 1mol of CaCO3 (100 g).
Amount of CaCO3 that willreact with 0.6844 g =
10073×0.6844= 0.9375 g
Question - 35 : - Chlorine is prepared in the laboratory bytreating manganese dioxide (MnO2) with aqueous hydrochloric acidaccording to the reaction
4HCl(aq) +MnO2(s) → 2H2O(l) +MnCl2(aq) + Cl2(g)
How many grams of HCl react with 5.0 g of manganesedioxide?
Answer - 35 : -
1 mol [55 + 2 × 16 = 87 g] MnO2 reactscompletely with 4 mol [4 × 36.5 = 146 g] of HCl.
5.0 g of MnO2 will reactwith
of HCl
= 8.4 g of HCl
Hence, 8.4 g of HCl will react completely with 5.0 g ofmanganese dioxide.