Solutions and Colligative Properties 2 Question 14
14. Pure water freezes at $273 \mathrm{~K}$ and 1 bar. The addition of 34.5 $\mathrm{g}$ of ethanol to $500 \mathrm{~g}$ of water changes the freezing point of the solution. Use the freezing point depression constant of water as $2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [Molecular weight of ethanol is $46 \mathrm{~g} \mathrm{~mol}^{-1}$ ]
(2017 Adv.)
Among the following, the option representing change in the freezing point is (a)
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Solution:
- $-\Delta T_{f}=i k_{f} m_{2}$
$$ =1 \times 2 \times \frac{34.5}{46 \times 500} \times 1000=3 $$
Vapour pressure curves shown in $(b)$ is in agreement with the calculated value of $-\Delta T_{f}$. (a) is wrong, vapour pressure decreases on cooling.
