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1.2 gm of sample chalk ( CaCO3 ) containing clay as impurity was treated with excess of dil HCl. The volume of CO2 liberated at STP was 230 ml. Calculate the percentage of impurity of the sample.

Given that , mass of CaCO₃ = 1.2 gm

                               Vol of CO₂ = 230 ml

CaCO₃ + HCl ? CaCl₂ + CO₂ + H₂0
(M : 100 )

22.4 l of CO₂ gms is liberated from 100 gms of CaCO₃.

.^.. 230 ml contains =

                                 =

                                 = 1. 03 gm of CaCO₃

1.2 gm of chalk contains 1.03 gm of pure CaCO₃

Percentage of purity =

                                           = 85.83 %

Percentage of purity    = 100 - 85.83

                                            = 14. 17 %