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

$20 \ g$ of naphthoic acid $(C_{11}H_8O_2)$ dissolved in $50 \ g$ of benzene $(K_f = 1.72 \ K \ kg \ mol^{-1})$ shows a depression in freezing point of $2 \ K$. The Van't Hoff factor is?

If a solute associates in a solvent, its experimentally calculated molar mass using the boiling point elevation method will be

The Van't Hoff factor calculated from association data is always ... than calculated from dissociation data.

If a solution prepared by adding $6.1 \ g$ of benzoic acid $(C_6H_5COOH)$ in $500 \ g$ of benzene freezes at $-0.290 \ ^oC$,find the percentage of association. (Given: $K_f$ of benzene $= 5.12 \ K \ kg/mol$,freezing point of pure benzene $= 5.50 \ ^oC$,molar mass of benzoic acid $= 122 \ g/mol$) (in $\%$)

When $0.0106 \text{ mole}$ of acetic acid is dissolved in $1 \text{ kg}$ of water, the observed freezing point depression is $0.0205 \text{ K}$. If the calculated freezing point depression is $0.0197 \text{ K}$, the Van't Hoff factor $(i)$ and the degree of dissociation $(\alpha)$ of acetic acid are respectively: