${\mathrm{K}}_2{\mathrm{HgI}}_4$ is 40% ionised in aqueous solution. The value of its van't Hoff factor $\left(\mathrm{i}\right)$ is:

Solute A associates in water. When $0.7 \mathrm{g}$ of solute $\mathrm{A}$ is dissolved in $42.0 \mathrm{g}$ of water, it depresses the freezing point by $0.2 °\mathrm{C}$. The percentage association of solute $\mathrm{A}$ in water, is

[Given: Molar mass of $\mathrm{A}=93 \mathrm{g}{ \mathrm{mol}}^{-1}$. Molal depression constant of water is $1.86 \mathrm{K} \mathrm{kg}{ \mathrm{mol}}^{-1}$]

Consider the following equilibrium for benzoic acid in benzene

$2{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}\rightleftharpoons {\left({\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}\right)}_2$

If $\mathrm{x}$ represents the degree of association of the solute then relationship between total number of moles of particles at equilibrium and van't Hoff factor $\left(\mathrm{i}\right)$ is

Determination of the molar mass of acetic acid in benzene using freezing point depression is affected by:

When $0.4 \mathrm{g}$ of acetic acid is dissolved in $40 \mathrm{g}$ of benzene, the freezing point of the solution is lowered by $0.45 \mathrm{K}$. Calculate the degree of association of acetic acid. Acetic acid forms dimer when dissolved in benzene. (${\mathrm{K}}_{\mathrm{f}}$ for benzene$=5.12 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}$, atomic mass: $\mathrm{C}=12, \mathrm{H}=1, \mathrm{O}=16$).