A $$0.5$$ percent solution of potassium chloride was found to freeze at $$-0.24°$$C. The percentage dissociation of potassium chloride is (Nearest integer)
(Molal depression constant for water is $$1.80$$ K kg mol$$^{-1}$$ and molar mass of KCl is $$74.6$$ g mol$$^{-1}$$)

We need to find the percentage dissociation of KCl given the freezing point depression data.

Calculate molality

A 0.5% solution means 0.5 g of KCl in 100 g of solution.

Mass of solvent (water) = 100 - 0.5 = 99.5 g = 0.0995 kg

Moles of KCl = $$\frac{0.5}{74.6}$$ = 0.006702 mol

Molality (m) = $$\frac{0.006702}{0.0995}$$ = 0.06736 mol/kg

Calculate the observed van't Hoff factor (i)

Using the formula: $$\Delta T_f = i \times K_f \times m$$

$$0.24 = i \times 1.80 \times 0.06736$$

$$0.24 = i \times 0.12125$$

$$i = \frac{0.24}{0.12125} = 1.979$$

Calculate percentage dissociation

For KCl dissociating as: $$KCl \to K^+ + Cl^-$$

Number of ions (n) = 2

The van't Hoff factor is related to degree of dissociation ($$\alpha$$) by:

$$i = 1 + \alpha(n - 1)$$

$$1.979 = 1 + \alpha(2 - 1)$$

$$1.979 = 1 + \alpha$$

$$\alpha = 0.979$$

Percentage dissociation = $$\alpha \times 100 = 97.9\% \approx 98\%$$

The correct answer is 98.