The molecular weight of benzoic acid in benzene as determined by the depression in freezing point method corresponds to:

$2\,g$ of benzoic acid $(C_6H_5COOH)$ dissolved in $25\,g$ of benzene shows a depression in freezing point equal to $1.62\,K$. The molal depression constant $(K_f)$ for benzene is $4.9\,K\,kg\,mol^{-1}$. What is the percentage degree of association of the acid?

If the degree of dissociation of an aqueous solution of a weak monobasic acid is determined to be $0.3$,then the observed freezing point depression will be $.....\%$ higher than the expected/theoretical freezing point depression. (Nearest integer)

$A$ solute $A$ dimerizes in water. The boiling point of a $2 \, M$ solution of $A$ is $100.52^{\circ} C$. The percentage association of $A$ is ..... $\%$.
(Round off to the Nearest integer)
[Use : $K_{b}$ for water $= 0.52 \, K \, kg \, mol^{-1}$,Boiling point of water $= 100^{\circ} C$]

The observed and normal masses of compound $MX_2$ are $65.6$ and $164$ respectively. The percent degree of ionisation of $MX_2$ is . . . . . . $\%.$ (Nearest integer)

Phenol associates in benzene to a certain extent to form a dimer. $A$ solution containing $20 \times 10^{-3} \ kg$ of phenol in $1 \ kg$ of benzene has its freezing point decreased by $0.69 \ K$. Calculate the percentage degree of association of phenol. ($K_f$ for benzene $= 5.12 \ K \ kg \ mol^{-1}$)