Question

Phenol associates in benzene to a certain extent to form a dimer. A solution containing $20\times {10}^{-3}$kg phenol in 1 kg of benzene has its freezing point depressed by 0.69 K. Calculate the fraction of phenol that has dimerised.

$K_f$for benzene $=5.12$ kg mol${}^{-1}K$

• A. $a=0.27$
• B. $a=0.38$
• C. $a=0.73$
• D. $a=0.62$

Answer 1

The correct option is C $a=0.73$

$\Delta T_f={T_f}^0-T_f=i.K_b.m$.

$\Delta T_f$ is freezing point depression.

$T_f$ is freezing point of solution.

$K_b$ is depression constant.

Molality $=m=\frac{\frac{20}{94}}{1}=0.212$

$\Delta T_f=0.69K$

$K_f=5.12$

we get, $i=\frac{0.69}{0.212\times 5.12}=0.635$

For phenol in benzene,

$2C_6{H}_{5}OH\leftrightharpoons {{(C}_6{H}_{5}OH)}_2$

If $x$ represents the degree of association of the solute, then we would have $(1–x)$ mol of benzoic acid left in unassociated form and correspondingly $2x$ as associated moles of benzoic acid at equilibrium. Therefore, total number of moles of particles at equilibrium is

$1-x+\frac{x}{2}=1-\frac{x}{2}=i$

$x=2(1-i)$

$⟹x=2(1-0.635)⟹x=0.733$

The fraction of phenol immersed will be $=x=0.733$

Hence, the correct option is $C$