KBr is 80% dissociated in an aqueous solution of 0.5 M concentration. (Given K_f for water = 1.86 K kg mol^–1). The solution freezes at:

271.326 K

The correct answer is option 1. 271.326 K.

We know,

\(\Delta T_f = K_f \cdot m \cdot i\)

where:

\(\Delta T_f\) is the depression in freezing point,

\(K_f\) is the cryoscopic constant of the solvent (water),

\(m\) is the molality of the solute (ethylene glycol), and

\(i\) is the Van't Hoff factor, which represents the number of particles formed when the solute dissolves.

Here, the degree of dissociation \(= 80\)%

Since, \(i = 1 − \alpha + n\alpha\) ⇌ \(K^+ + Br^-\)

\(∴ i = \frac{1 − 0.8 + 2 × 0.8}{1} =1.8\)

So, \(\Delta T_f = 1.8 × 1.86 × 0.5\)

\(⇒ \Delta T_f = 1.67 k\)

Depression in freezing point,

\(\Delta T_f = T_f^0 − T_f \)

\(1.674 = 273 − T_f\)

\(T_f = 273 − 1.674 = 271.326 K \)