Question

1g of an impure sample of calcium carbonate(containing
no thermally decomposable impurities) on complete
thermal decomposition gave 0.33 g of carbon dioxide
gas. The percentage of impurity in this sample is(Atomic
masses of C, O and Ca are respectively 12, 16 and
40 g mol)
(a) 25
(b) 20
(c) 30
(d) 33 ​

Answer 1

Percentage of impurities = 25%

Explanation:

CaCO3 = CaO + CO2

Molar mass of CaCO3 = 40 + 12 + 16*3 = 100

Molar mass of CO2 = 12 + 16*2 = 44

44g of CO2 is obtained from 100g of CaCO3

Given that 0.33 g of CO2 is obtained from = (0.33 * 100) / 44

                                                     = 0.75g of CaCO3

Mass of impurities = 1 - 0.75 = 0.25

Percentage of impurities = (0.25/1) * 100  

                                       = 25%

Option A is the answer.