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 $K_{f} = 1.86 \ K \ kg \ mol^{-1}$)

$A$ non-volatile solute '$A$' tetramerizes in water to the extent of $80\%$. $2.5 \text{ g}$ of '$A$' in $100 \text{ g}$ of water lowers the freezing point by $0.3^\circ \text{C}$. The molar mass of $A$ in $\text{g mol}^{-1}$ is ($K_f$ for water $= 1.86 \text{ K kg mol}^{-1}$)

Molecules of benzoic acid $(C_6H_5COOH)$ dimerise in benzene. '$w$' $g$ of the acid dissolved in $30 \ g$ of benzene shows a depression in freezing point equal to $2 \ K.$ If the percentage association of the acid to form dimer in the solution is $80,$ then $w$ is : ............. $g$
(Given that $K_f = 5 \ K \ kg \ mol^{-1},$ Molar mass of benzoic acid $= 122 \ g \ mol^{-1}$ )

The degree of dissociation $(\alpha)$ of a weak electrolyte,$A_xB_y$ is related to van't Hoff factor $(i)$ by the expression

The values of observed and calculated molecular weight of silver nitrate are $90$ and $170$ respectively. The degree of dissociation of silver nitrate is ......... $\%$.

Three particles of a solute $A$ associate in benzene to form a trimer $A_3$. Calculate the freezing point of a $0.25 \ m$ solution if the degree of association of solute $A$ is $0.80$. The freezing point of benzene is $5.5 \ ^oC$ and $K_f$ is $5.12 \ K \ kg \ mol^{-1}$. (Answer in $K$)