Question

Pure water freezes at 273 K and 1 bar. The addition of 34.5g of ethanol to 500g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2K kg $mo{l}^{-1}$. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [Molecular weight of ethanol is 46g $mo{l}^{-1}$]

Among the following, the option representing change in the freezing point is

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

Step 1:

The number of moles of ethanol n = $\frac{34.5g}{46g{mol}^{-1}}=0.75mol$

Molality m = $\frac{0.75mol}{0.5Kg}=1.5m$

Step 2:

The depression in freezing point $\Delta T_f=K_f^m=2KKg{mol}^{-1}\times 1.5m=3K$

The freezing point of pure water is 273 K.

Freezing point of solution = 273 − 3 = 270 K.

Option (A) represents the correct graph. In this graph, the freezing point of the solution is shown as 270 K. Also with the increase in temperature, the vapour pressure of pure water and mixture increases.

Hence, option (A) is correct.