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)

75

Answer 2

$\underset{1 mol}{CaCO _{3}} \stackrel{\Delta}{\longrightarrow} \underset{1 mol}{CaO} + CO _{2}$
$\underset{1mol}{CaCl _{2}}+ Na _{2}CO _{3} \longrightarrow \underset{1mol}{CaCO _{3}}+2 Na$
$1 mol CaO \cong 1 mol CaCl _{2}$
$\frac{0.56}{56} mol CaO \cong 0.01 mol CaCl _{2}$
$=0.01 \times 111\, g CaCl _{2}$
$=1.11\, g CaCl _{2}$

Thus, in the mixture, weight of

$NaCl =4.44- 1.11=3.33\, g$
$\therefore$ Percentage of $NaCl =\frac{3.33}{4.44} \times 100$
$=75 \%$