Question

A mixture of $CaCl_2$ and $NaCl$ weighing $4.44\, g$ is treated with sodium carbonate solution to precipitate all the calcium 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 $Ca = 40)$

• A. 30.6
• B. 75
• C. 69.4
• D. 25

Answer 1

Answer: (B)

According to question,
$\underset{1\,mol}{CaCl _{2}}+ Na _{2} CO _{3}+ NaCl \longrightarrow \underset{1\,mol}{CaCO _{3}}$
$\underset{1\,mol}{CaCO _{3}} \xrightarrow{\Delta} \underset{1 mol }{ CaO }+ CO _{2}$
$1 mol$ of $CaO \cong 1 mol$ of $CaCl _{2}$
We know that, number of moles
$=\frac{\text { Mass }}{\text { Molecular mass }}$
$\frac{0.56}{56} mol$ of $CaO \cong = 0.01\, mol$ of $CaCl _{2}$
$=0.01 \times 111\, g\, CaCl _{2}$
$=1.11\, g\, CaCl _{2}$
Thus, in the mixture, weight of $NaCl$
$=4.44-1.11=3.33\, g$
$\therefore$ Percentage of $NaCl =\frac{3.33}{4.44} \times 100=75 \%$