If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution,…

If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water $\left(\Delta T_{f}ight)$, when $0.01 \mathrm{~mol}$ of sodium sulphate is dissolved in $1 \mathrm{~kg}$ of water, is $\left(K_{f}=1.86ight.$ $\left.\mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}ight)$
  1. $0.372 \mathrm{~K}$
  2. $0.0558 \mathrm{~K}$
  3. $0.0744 \mathrm{~K}$
  4. $0.0186 \mathrm{~K}$

Solution

Sodium sulphate dissociates as $\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{~s}) \longrightarrow 2 \mathrm{Na}^{+}+\mathrm{SO}_{4}^{2-}$
hence van't hoff factor $i=3$
Now $\Delta T_{f}=i k_{f} \cdot m$
$=3 \times 1.86 \times 0.01=0.0558 \mathrm{~K}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

Practice more SOLUTIONS questions on Aicharya