Question

The boiling point of an aqueous solution of a non-volatile solute is $100.15{}^{\circ }C$. What is the freezing point of an aqueous solution obtained by diluting the above solution with an equal volume of water?
The value of $K_b$ and $K_f$ for water are $0.512{}^{\circ }\text{C kg/mol}$ and $1.86{}^{\circ }\text{C kg/mol}$ respectively.

• A. $-0.545{}^{\circ }C$
• B. $-0.520{}^{\circ }C$
• C. $-0.273{}^{\circ }C$
• D. $-1.86{}^{\circ }C$

Answer 1

The correct option is C $-0.273{}^{\circ }C$

$\frac{△T_f}{△T_b}=\frac{K_f}{K_b}$

$\Rightarrow △T_f=\frac{K_f}{K_b}\times △T_b$

$\therefore △T_f=\frac{0.15\times 1.86}{0.512}=0.545{}^{\circ }C$

On diluting the solution to double,
$△T_f\propto \frac{1}{\text{solvent weight}}$

or, $△T_f=\frac{0.545}{2}=0.2725{}^{\circ }C$

$T_f=T_f^{\circ }-0.273{}^{\circ }C=-0.273{}^{\circ }C$