Solutions and Colligative Properties 2 Question 4
4. Molal depression constant for a solvent is $4.0 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. The depression in the freezing point of the solvent for 0.03 mol $\mathrm{kg}^{-1}$ solution of $\mathrm{K}{2} \mathrm{SO}{4}$ is
(Assume complete dissociation of the electrolyte)
(2019 Main, 9 April II)
(a) $0.18 \mathrm{~K}$
(b) $0.36 \mathrm{~K}$
(c) $0.12 \mathrm{~K}$
(d) $0.24 \mathrm{~K}$
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Solution:
Key Idea Depression in freezing point $\left(\Delta T_{f}\right)$ is given by $\Delta T_{f}=i K_{f} m$
$i=$ vant Hoff factor
$K_{f}=$ molal depression constant $m=$ molality
$$ \begin{aligned} K_{f} & =4.0 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \ m & =0.03 \mathrm{~mol} \mathrm{~kg}^{-1} \ \Delta T_{f} & =? \end{aligned} $$
For $\mathrm{K}{2} \mathrm{SO}{4}, i=3$
It can be verified by the following equation :
$$ \mathrm{K}{2} \mathrm{SO}{4} \rightleftharpoons 2 \mathrm{~K}^{+}+\mathrm{SO}_{4}^{2-} $$
Using formula
$$ \begin{aligned} \Delta T_{f} & =i K_{f} \times m \ \Delta T_{f} & =3 \times 4 \times 0.03=0.36 \mathrm{~K} \end{aligned} $$
