Question

The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g of benzene is lowered by ${0.45}^{\circ }C.$ Calculate the degree of association of acetic acid in benzene :

$(K_f$ for benzene $=5.12Kmo{l}^{-1}kg)$

• A. $5.5$%
• B. $64.5$%
• C. $35.5$%
• D. $94.6$%

Answer 1

The correct option is D $94.6$%

Freezing point depression is a colligative property observed in solutions that result from the introduction of solute molecules to a solvent.

The freezing points of solutions are all lower than that of the pure solvent and is directly proportional to the molality of the solute.

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

$\Delta T_f$ is freezing point depression.

$T_f$ is freezing point of solution.

$K_b$ is depression constant.

Molality $=m=\frac{\frac{0.2}{60}}{\frac{20}{1000}}=0.166$

$\Delta T_f=0.45K$

$K_f=5.12$

we get, $i=\frac{0.45}{0.1666\times 5.12}=0.527$

For acetic acid in benzene, dimerisation takes place :

$2C{H}_{3}COOH\leftrightharpoons {\left(C{H}_{3}COOH\right)}_2$

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 $x/2$ as associated moles of benzoic acid at equilibrium.

Therefore, the total number of moles of particles at equilibrium is

$i=1-x+\frac{x}{2}⟹x=2(1-i)⟹x=0.946$

The percentage of the association will be $94.6\%$.

Therefore, option is D.