Solutions and Colligative Properties - Result Question 14-1
14. Pure water freezes at $273$ K and $1$ bar. The addition of $34.5$ g of ethanol to $500$ 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
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Answer:
Correct Answer: 14. (b)
Solution:
$ \begin{aligned} -\Delta T_f & =i k_f m_2 \\ & =1 \times 2 \times \frac{34.5}{46 \times 500} \times 1000=3 \end{aligned} $
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.
BETA
