When $20 \ g$ of naphthoic acid $(C_{11}H_8O_2)$ is dissolved in $50 \ g$ of benzene, a freezing point depression of $2 \ K$ is observed. The van't Hoff factor $(i)$ is [$K_f = 1.72 \ K \ kg \ mol^{-1}$].

$2$ molal solution of a weak acid $HA$ has a freezing point of $-3.885^{\circ} C$. The degree of dissociation of this acid is ........ $\times 10^{-3}$. (Round off to the Nearest Integer).
[Given: Molal depression constant of water = $1.85 \ K \ kg \ mol^{-1}$,Freezing point of pure water = $0^{\circ} C$]

If the degree of association is $70 \%$ for the reaction $2 A \rightleftharpoons (A)_2$, the van't-Hoff factor for the solute $A$ is

$A$ compound $MX_2$ has observed and normal molecular masses $65.6$ and $164$ respectively. Calculate the percentage of ionization of $MX_2$ ............ $\%$

Benzoic acid undergoes dimerisation in benzene solution. The van't Hoff factor $(i)$ is related to the degree of association '$x$' of the acid as

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)