Question

$20g$ of solvent containing $0.25g$ of solute (mol. mass 122 g/mol) freezes at ${0.3}^{\circ }C$ below the freezing point of pure solvent. If the solvent exists as dimer in the solvent, find its degree of association. $(K_f{)}_{\text{solvent}}=5.12Kk{g}^{-1}mo{l}^{-1}$

• A. $0.01$
• B. $0.99$
• C. $0.14$
• D. $0.86$

Answer 1

The correct option is D $0.86$

$\text{Weight of solute}\left(w\right)=0.25g\text{Weight of solvent}\left(W\right)=20g=\frac{20}{1000}kg\Delta T_f={0.3}^{\circ }C=0.3K$

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

$m_{obs}=\frac{K_f\times w}{W\times \Delta T}{m}_{obs}=\frac{5.12\times 0.25\times 1000}{20\times 0.3}=213.3gmo{l}^{-1}$

$\text{Theoretical molecular mass of solute}=122gmo{l}^{-1}$
So, $i=\frac{\text{Theoretical molecular mass}}{\text{Observed molecular mass}}=\frac{122}{213.3}=0.57$

Degree of association $\left(\beta \right)$ :
$\beta =\frac{(i-1)\times n}{(1-n)}$
Since solute dimerises, so $n=2$
Putting the values:
$\beta =\frac{(0.57-1)\times 2}{(1-2)}\beta =0.86$