20g of a binary electrolyte (Molecular mass = 100) are dissolved in 500 g of water. The freezing point of the solution is \(-0.74^oC\) and \(k_f = 1.86 \text{ K kg mol}^{-1}\). The degree of ionization of electrolyte is:

0.0

The correct answer is option 4. 0.0.

We know from the Depression of the freezing point,

\(\Delta T_f = iK_f × m\)

Where, \(\Delta T_f\) is the depression in freezing point, \(K_f\) is ebullioscopic constant,

\(m\) is the molality, \(i\) is the van't Hoff factor

Applying the values in the equation we get:

\(0.744 = i × 1.86 × \frac{20}{100} × \frac{1000}{500}\)

or, \(i = \frac{5 × 0.744}{1.86 × 2} = 1\)

Also,

\(\alpha = \frac{1 - i}{n - 1} = \frac{1 - 1}{2 - 1} = \frac{0}{1} = 0\)