A mixture of \(\mathrm{CaCl}_{2}\) and \(\mathrm{NaCl}\) weighing \(4.44 \mathrm{~g}\) 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 \(0.56 \mathrm{~g}\) of \(\mathrm{CaO}\). The percentage of \(\mathrm{NaCl}\) in the mixture (atomic mass of \(\mathrm{Ca}=40\) ) is

\(\underset{1 \mathrm{~mol}}{\mathrm{CaCO}_{3}} \stackrel{\Delta}{\longrightarrow} \underset{1 \mathrm{~mol}}{\mathrm{CaO}}+\mathrm{CO}_{2}\)
\(\mathrm{CaCl}_{2}+\mathrm{Na}_{2} \mathrm{CO}_{3} \longrightarrow \underset{1 \mathrm{~mol}}{\mathrm{CaCO}_{3}}+2 \mathrm{Na}\)
\(1 \mathrm{~mol} \mathrm{} \mathrm{CaO} \cong 1 \mathrm{~mol} \mathrm{} \mathrm{CaCl}_{2}\)
\(\frac{0.56}{56} \mathrm{~mol} \mathrm{CaO} \cong 0.01 \mathrm{~mol} \mathrm{CaCl}_{2}\)
\(=0.01 \times 111 \mathrm{~g} \mathrm{CaCl}_{2}\)
\(=1.11 \mathrm{~g} \mathrm{CaCl}_{2}\)
Thus, in the mixture, weight of
\(\mathrm{NaCl}=4.44-1.11=3.33 \mathrm{~g}\)
\(\therefore\) Percentage of \(\mathrm{NaCl}=\frac{3.33}{4.44} \times 100\)
\(=75 \%\)