A mixture of ${\mathrm{CaCl}}_2$ and NaCl weighing 4.44 g is treated with sodium carbonate solution to precipitate all the ${\mathrm{Ca}}^{2+}$ 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 is (Atomic mass of Ca = 40)

${\mathrm{CaCO}}_3\stackrel{\mathrm{\Delta }}{⟶}\mathrm{CaO}+{\mathrm{CO}}_2$

 1 mol          1 mol

$\underset{1 \mathrm{mol}}{{\mathrm{CaCl}}_2}+{\mathrm{Na}}_2{\mathrm{CO}}_3⟶\underset{1 \mathrm{mol}}{{\mathrm{CaCO}}_3}+2\mathrm{NaCl}$

$1\text{ mol of }\mathrm{CaO}\cong 1 \mathrm{mol}\text{ of }{\mathrm{CaCl}}_2$

$\frac{0.56}{56} \mathrm{mol}\text{ of }\mathrm{CaO}\cong 0.01 \mathrm{mol}\text{ of }{\mathrm{CaCl}}_2$

                            $=0.01\times 111 g {\mathrm{CaCl}}_2$

                             $=1.11 g {\mathrm{CaCl}}_2$

Thus, in the mixture, weight of NaCl = 4.44- 1.11= 3.33 g

$\therefore \text{ Percentage of }\mathrm{NaCl}=\frac{3.33}{4.44}\times 100=75\%$