Question

2.0 g of bensoic acid dissolved in 25.0g of benzene shows a depression in freezing point to 1.62K. Mollal depression constant ($K_f$ of bezene is 4.9 K.kg. $mo{l}^{-1}$. What is the percentage association of the acid?

• A. $\left(\alpha \right)$= 90.8%
• B. $\left(\alpha \right)$= 99.2%
• C. $\left(\alpha \right)$= 9.8%
• D. $\left(\alpha \right)$= 49.6%

Answer 1

The correct option is B $\left(\alpha \right)$= 99.2%

Freezing point depression is a colligative property observed in solutions that result from the introduction of solute molecules to a solvent. The freezing point 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 $=m=\frac{\frac{2}{122}}{\frac{25}{1000}}=0.655$

$\Delta T_f=1.62K$

$K_f=4.9$

We get, $i=\frac{1.62}{0.655\times 4.9}=0.504$

For benzoic acid in benzene,

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

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

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

$x=2(1-i)$

$⟹x=2(1-0.504)⟹x=0.992$

Percentage association is $99.2\%$.