A mixture of CaCl₂ and NaCl, weighing 4.44g is treated with sodium carbonate solution to precipitate all the Ca²⁺ ions as calcium carbonate. The calcium carbonate so obtained, is heated strongly to get 0.56 g of CaO. The percentage of NaCl in the mixture (Atomic mass of Ca = 40) is

Concerned reaction is
$\underset{100\mathrm{g}}{{\mathrm{CaCO}}_3}\stackrel{\mathrm{\Delta }}{⟶}\underset{56\mathrm{g}}{\mathrm{CaO}}+{\mathrm{CO}}_2$
56g of CaO is obtained from 100g of CaCO₃
$0.56\mathrm{g}\text{ of }\mathrm{CaO}\text{ is obtained by }\frac{100}{56}\times 0.56=1\mathrm{g}\text{ of }$
CaCO₃ we know that,
$\underset{111 \mathrm{g}}{{\mathrm{CaCl}}_2}+{\mathrm{Na}}_2{\mathrm{CO}}_3⟶\underset{100 \mathrm{g}}{{\mathrm{CaCO}}_3}+2\mathrm{NaCl}$
$100\mathrm{g}\text{ of }{\mathrm{CaCO}}_3\text{ is obtained by }111\mathrm{g}\text{ of }{\mathrm{CaCl}}_2$
$1\mathrm{g}\text{ of }{\mathrm{CaCO}}_3\text{ is obtained by }\frac{111}{100}=1.11\mathrm{g}\text{ of }{\mathrm{CaCl}}_2$
Weight of NaCl = 4.44 - 1.11 = 3.33 g
$\mathrm{\%}\text{ age of }\mathrm{NaCl}=\frac{3.33}{4.44}\times 100=75 \mathrm{\%}$