$0.2$ molal aqueous solution of $KCl$ freezes at $-0.680^{\circ} C$. Calculate van't Hoff factor for this solution. $(K_{f} = 1.86 \ K \ kg \ mol^{-1})$

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

In a solvent,$50\,\%$ of an acid $HA$ dimerizes and the rest dissociates. The van't Hoff factor of the acid is $.....\times 10^{-2}$. (Round off to the nearest integer)

Determine the degree of association (polymerization) for the reaction in aqueous solution $6 HCHO \rightleftharpoons C_6H_{12}O_6$. If the observed (mean) molar mass of the mixture is $150 \ g/mol$. (Given: Molar mass of $HCHO = 30 \ g/mol$)

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,$K_b = 2.65 \ K \ m^{-1}$,$K_f = 5.12 \ K \ m^{-1}$,$T_b^o = 80 \ ^\circ C$,$T_f^o = 5.5 \ ^\circ C$)

If $0.15 \ m$ aqueous solution of $KCl$ freezes at $-0.51^{\circ} C$,calculate the van't Hoff factor of $KCl$ (cryoscopic constant of water is $1.86 \ K \ kg \ mol^{-1}$).