Question

A mixture of $CaC{l}_2$ and $NaCl$ weighing $4.44g$ 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 $0.56g$ 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,
$\underset{1mol}{CaC{l}_2}+N{a}_{2}C{O}_3+NaCl⟶\underset{1mol}{CaC{O}_3}$
$\underset{1mol}{CaC{O}_3}\stackrel{\Delta }{\to }\underset{1mol}{\mathrm{CaO}}+C{O}_2$
$1mol$ of $CaO\cong 1mol$ of $CaC{l}_2$
We know that, number of moles
$=\frac{\text{ Mass }}{\text{ Molecular mass }}$
$\frac{0.56}{56}mol$ of $CaO\cong =0.01mol$ of $CaC{l}_2$
$=0.01\times 111gCaC{l}_2$
$=1.11gCaC{l}_2$
Thus, in the mixture, weight of $NaCl$
$=4.44-1.11=3.33g$
$\therefore$ Percentage of $NaCl=\frac{3.33}{4.44}\times 100=75\%$