Question

Phenol associates in benzene to a certain extent for a dimer. A solution containing $20\times {10}^{-3}kg$ of phenol in $1kg$ of benzene has its freezing point decreased by $0.7K$
Calculate the fraction of the phenol that has dimerised
Given :
$(K_f{)}_{\text{Benzene}}=5.12Kkgmo{l}^{-1}$

• A. $0.29$
• B. $0.91$
• C. $0.09$
• D. $0.72$

Answer 1

The correct option is D $0.72$

$\text{Weight of phenol}\left(w\right)=20\times {10}^{-3}kg=20g\text{Weight of benzene}\left(W\right)=1kg\Delta T_f=0.7K$

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

$m_{obs}=\frac{K_f\times w}{W\times \Delta T}{m}_{obs}=\frac{5.12\times 20}{1\times 0.7}=146.3gmo{l}^{-1}$
$\text{Theoretical molecular mass of phenol}\left(C_6{H}_{5}OH\right)=94gmo{l}^{-1}$
So, $i=\frac{\text{Theoretical molecular mass}}{\text{Observed molecular mass}}=\frac{94}{146.3}=0.64$

Degree of association $\left(\beta \right)$ :
$\beta =\frac{(i-1)\times n}{(1-n)}$
Since phenol dimerises, so $n=2$
Putting the values:
$\beta =\frac{(0.64-1)\times 2}{(1-2)}\beta =0.72$