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