If a $0.01 \ m$ aqueous solution of an electrolyte freezes at $-0.056 \ ^{\circ}C$,calculate the van't Hoff factor $(i)$ for the electrolyte. (Cryoscopic constant of water $K_{f} = 1.86 \ K \ kg \ mol^{-1}$)

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

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)

The difference between the boiling point and the freezing point of a $0.2 \ m$ solution of acetic acid in benzene is $75.7 \ ^\circ C$. Calculate the value of the van't Hoff factor $i$. (For benzene,$K_b = 2.65 \ K \ m^{-1}$,$K_f = 5.12 \ K \ m^{-1}$,$T_b^o = 80 \ ^\circ C$,$T_f^o = 5.5 \ ^\circ C$)

$A$ solution of $1.25 \ g$ of '$P$' in $50 \ g$ of water lowers the freezing point by $0.3^{\circ}C$. The molar mass of '$P$' is $94 \ g \ mol^{-1}$. Given $K_{f(\text{water})} = 1.86 \ K \ kg \ mol^{-1}$,the degree of association of '$P$' in water is:

In a solution with molality $m$,if the solute exists as a trimer,what would be the depression in the freezing point of the solution?