28 g KOH is required to completely neutralise `CO_2` produced on heating 60 g of impure `CaCO_3` The percentage purity of `CaCO_3` is approximately (molar masses of KOH and `CaCO_3` are 56 and `100 g "mol"^(-1)` , respectively)

Options

A. 41.6
B. 40
C. 20.8
D. 83.3

Answer

A

Explanation

The given relation `CaCO_3 underset(Delta)overset(2KOH)(to) CO_2` means , 2 moles of KOH will neutralise 1 mole of `CO_2`
Given ,
(i) KOH (used ) = 28 g
(ii) Molar mass of KOH (M) = 56 g
(iii) Molar mass of `CaCO_3 = 60 g`
`because 112 g (56 xx 2) `of KOH will neutralise
= 100 g or `CaCO_3`
` therefore 28g` KOH will neutralise ` = (100 xx 28)/(112) = 25g` of `CaCO_3`
Also ,
` because ` 60 g (impure ) of `CaCO_3` has 25g pure `CaCO_3`
`therefore 100g` (impure) `CaCO_3` has pure `CaCO_3`
` =(25 xx 100)/(60) = 41.6 g`