28g 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

Given, Wight of used $KOH=28g$

Weight of impure ${CaCO}_3$ = $60g$

${CaCO}_3\underset{\Delta }{\stackrel{2KOH}{⟶}}{CO}_2$

 According to the given reaction 2 moles of KOH is used to neutralise 1 mole of ${CaCO}_3$

Molar mass of NaOH = 56g

Molar mass of ${CaCO}_3$ = 100g

112g of NaOH neutralise = 100g of ${CaCO}_3$

28g of NaOH neutralise = $\frac{28 }{112}\times 100$= 25g of ${CaCO}_3$

Pure ${CaCO}_3$ in 60g of impure ${CaCO}_3$ = 25g

Pure ${CaCO}_3$ in 100g of ${CaCO}_3$ = $25\times \frac{100}{60}$ = 41.6g 

Therefore, option A is correct.