Practicing Success
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: |
0.5 0.75 1.0 0.0 |
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, \(\alpha = \frac{1 - i}{n - 1} = \frac{1 - 1}{2 - 1} = \frac{0}{1} = 0\) |