$10 g$ of a mixture of ${CaCO}_3$ and ${Na}_2{CO}_3$ on ignition suffered a loss in weight of $2.2 g$. The mass ratio of ${CaCO}_3$ and ${Na}_2{CO}_3$ is

 Let the weight of ${CaCO}_3$ be x g

So, weight of ${Na}_2{CO}_3$ = (10-x) g

Moles of ${CaCO}_3=\frac{x}{100}$

Balanced chemical reaction

${CaCO}_3\to CaO+{CO}_2$

Moles of ${CO}_2$ formed = $x/100$

$\frac{x}{100}\times 44=2.2$

$x=2.2\times \frac{100}{44}$

$x=5$

Weight of ${CaCO}_3$ = 5g.

Weight of ${Na}_2{CO}_3$ = 10 –5 = 5g.

So, the mass ratio of ${CaCO}_3$ and ${Na}_2{CO}_3$ is 1:1.

Therefore, option A is correct.