van't Hoff factor of a 0.5% (w/w) aqueous solution of KCl which freezes at $-0.{24}^0\mathrm{C}$ is $({\mathrm{K}}_f$ of water =1.86K ${\mathrm{kgmol}}^{-1}$, mol wt. of KCl=74.5)

Given data : It is given to identifyVan't Hoff factor of an aqueous KCl solution at 0.5% (w/w), which freezes at$-0.{24}^0\mathrm{C}$ is $({\mathrm{K}}_f$ of water =1.86K ${\mathrm{kgmol}}^{-1}$mol weight of KCl = 74.5)

Explanation

The Van't Hoff factor sheds light on how solutes affect the collaberative characteristics of solutions. The letter I is used to represent it. The ratio of a substance's mass concentration to the concentration of the particles that are produced when it dissolves is known as the Van't Hoff factor.

The Van't Hoff factor describes how much a substance associates or dissociates in a solution. For instance, the value of I is typically 1 when a non-electrolytic substance is dissolved in water. The value of I however, is equal to the total number of ions present in one formula unit of the substance when an ionic compound forms a solution in water.

$∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{iK}}_{\mathrm{f}}\mathrm{m}$

$0.24=\mathrm{i}\times 1.86\times \frac{0.5}{74.5}\times \frac{1000}{100}$

i=1.92

Hence, the correct option (C). 1.92