Question

$5 \,g$ of $Na_2SO_4$ was dissolved in $x\, g$ of $H_2O$. The change in freezing point was found to be $3.82^{\circ}C$. If $Na_2SO_4$ is $81.5\%$ ionised, the value of $x$
$(K_f$ for water = $1.86^{\circ}C \, kg \, mol^{-1})$ is approximately :
(molar mass of $S = 32 \, g \, mol^{-1}$ and that of $Na = 23 \, g \, mol^{-1}$)

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

Answer 1

Answer: (C)

$Na_{2}SO_{4} \rightarrow 2Na^{+}+SO_{4}^{2-}$
$x = 1 + \left(3 - 1\right) 0.815 = 2.63$
$3.82 = 1.86 × 2.63\times\frac{5\times 1000}{142\times x}$
$\therefore x=\frac{1.86\times 2.63\times 5000}{142\times 3.82}$
$=45\, gm$