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:
Answer & explanation
Correct answer: option 4
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,
\(\alpha = \frac{1 - i}{n - 1} = \frac{1 - 1}{2 - 1} = \frac{0}{1} = 0\)