Question

An aqueous solution freezes at 272.4 K while pure water freezes at 273 K. Determine the boiling point of the solution.

[Given : $K_f=1.86$ K kg mol${}^{-1}$, $K_b=0.512$ K kg mol${}^{-1}$ and vapour pressure of water at $298K=23.756$ mm of Hg].

• A. $300.73$ K
• B. $373.165$ K
• C. $400$ K
• D. $273.15$ K

Answer 1

The correct option is B $373.165$ K

Given, the freezing point of pour water $=273K$

The freezing point of the solution $=272.4K$

Depression in the freezing point $\Delta T_f=273-272.4$

$K_f=1.86Kg/mol$

$\Delta T_b=K_b\times m$

Molality (m) $=\frac{\Delta T_f}{K_f}=\frac{0.6}{1.86}=0.322m$

$\Delta T_b=0.512\times 0.322$

$=0.164K$

Boiling point of the solution $=373+0.164$

$=373.164K$

Option B is correct.