The relationship between the van’t Hoff factor $(i)$ and the degree of dissociation $(\alpha)$ is:

$0.8 \ mL$ of acetic acid of density $1.06 \ g \ mL^{-1}$ when dissolved in $1 \ kg$ of water causes a depression in freezing point by $0.0325^{\circ} C$. The Van't Hoff factor is:
$(K_{f} \text{ of } H_2O = 1.86 \ K \ kg \ mol^{-1})$

$A$ $0.0020 \ m$ aqueous solution of an ionic compound $[Co(NH_3)_5(NO_2)]Cl$ freezes at $-0.00732 \ ^oC.$ Number of moles of ions which $1 \ mol$ of ionic compound produces on being dissolved in water will be
$(K_f = 1.86 \ ^oC/m)$

If $1$ mole of a solute dissociates to give $n$ number of ions,then the degree of dissociation $\alpha$ in terms of the van't Hoff factor $i$ is given by:

$A$ $0.2 \ m$ solution of acid $HX$ is $20\%$ ionized in the solution. Given $K_f = 1.86 \ K \ kg \ mol^{-1}$,the freezing point of the solution is ........... $^oC$.

Calculate the molality of an aqueous solution of an electrolyte that freezes at $-0.93 \text{ }^\circ\text{C}$. Given that $K_f$ for water is $1.86 \text{ K kg mol}^{-1}$ and the van't Hoff factor $(i)$ is $1.25$. (Freezing point of pure water $= 0 \text{ }^\circ\text{C}$) (in $\text{ m}$)