Question

A solution which has $10g$ of solvent and $0.20g$ of solute (mol. mass 122 g/mol) freezes at ${0.5}^{\circ }C$ below the freezing point of pure solvent. If the solute exists as dimer in the solvent, choose the correct option :
$(K_f{)}_{\text{solvent}}=5.12Kk{g}^{-1}mo{l}^{-1}$
Here,
$i=\text{Van't hoff factor}\beta =\text{Degree of association}$

• A. $\beta =0.94$
• B. $i=0.59$
• C. $\beta =0.51$
• D. $i=0.75$

Answer 1

The correct option is B $i=0.59$

$\text{Weight of solute}\left(w\right)=0.20g\text{Weight of solvent}\left(W\right)=10g=\frac{10}{1000}kg\Delta T_f={0.5}^{\circ }C=0.5K$

$\text{Let}{m}_{obs}\text{be observed molecular mass of solute}$

$m_{obs}=\frac{K_f\times w}{W\times \Delta T}{m}_{obs}=\frac{5.12\times 0.20\times 1000}{10\times 0.5}=204.8gmo{l}^{-1}$

$\text{Theoretical molecular mass of solute}=122gmo{l}^{-1}$
So, $i=\frac{\text{Theoretical molecular mass}}{\text{Observed molecular mass}}=\frac{122}{204.8}=0.595$

Degree of association $\left(\beta \right)$ :
$\beta =\frac{(i-1)\times n}{(1-n)}$
Since solute dimerises, so $n=2$
Putting the values:
$\beta =\frac{(0.595-1)\times 2}{(1-2)}\beta =0.81$