Question

$5g$ of $N{a}_{2}S{O}_4$ was dissolved in $xg$ of $H_{2}O$. The change in freezing point was found to be ${3.82}^{o}C$. If $N{a}_{2}S{O}_4$ is $81.5\%$ ionised, the value of $x$
$(k_f$ for water $={1.86}^{o}Ckgmo{l}^{-1})$ is approximately:
(molar mass of $S=32gmo{l}^{-1}$ and that of $Na=23gmo{l}^{-1}$)

• A. $45g$
• B. $65g$
• C. $15g$
• D. $25g$

Answer 1

Answer: (A)

$45g$

Molar mass of $N{a}_{2}S{O}_4=2\left(23\right)+32+4\left(16\right)=142$ g/mol

5 g $N{a}_{2}S{O}_4=\frac{5g}{142g/mol}=0.0352$ mol

Mass of water $=x$ g $=\frac{x}{1000}$ kg = $0.001x$ kg

Molality $m=\frac{\text{0.0352 mol}N{a}_{2}S{O}_4}{\text{0.001 x mol}{H}_{2}O}=\frac{35.2}{x}mol/kg$
$N{a}_{2}S{O}_4$ is $81.5$ % ionized.

The degree of dissociation $\alpha =\frac{81.5}{100}=0.815$

The vant Hoff factor $i=[1+(n-1\left)\alpha \right]$

$i=[1+(3-1)\times 0.815]=2.63$

The depression in the freezing point $\Delta T_f=i{K}_{f}m$

$3.82{}^{o}C=2.63\times 1.86{}^{o}CKg/mol\times \frac{35.2}{x}mol/kg$

$x=45$ g