Question

Pure water freezes at $273\text{K}$ and $1\text{bar}$. The addition of $34.5\text{g}$ of ethanol to $500\text{g}$ of water changes the freezing point of the solution. Use the freezing point depression constant of water as $2{\text{K kg mol}}^{-1}$. The figures shown below represent plots of vapour pressure (V.P.) versus temperature $\text{(T)}$. (Molecular weight of ethanol is $46{\text{g mol}}^{-1}$). Among the following, the option representing change in the freezing point is:

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Depression in freezing point of ethanol and water solution:
$\Delta T_f=i{K}_{f}m$
$=1\times 2\times \frac{34.5/46}{500/1000}=3K$
So the freezing point of the solution is 270 K

• In option b and d, the freezing point shown is 271 K so both are incorrect
• Between option a and c the vapour pressure of the solution is decreasing with temperature after the freezing point in option a so it is also incorrect while option c seems to be the only possible curve for the given conditions.