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
The depression of freezing point gives the van’t Hoff factor, and for a binary electrolyte AB → A+ + B−, i = 1 + α.
\( \displaystyle m = \frac{20/100\ \text{mol}}{0.5\ \text{kg}} = 0.4\ \mathrm{mol\,kg^{-1}} \)
Expected \( \Delta T_f \) (no ionisation) = \( K_f \times m = 1.86 \times 0.4 = 0.744\ \text{K} \)
Observed \( \Delta T_f = 0 - (-0.74) = 0.74\ \text{K} \)
\( \displaystyle i = \frac{0.74}{0.744} \approx 1.0 \)
\( \alpha = i - 1 = 1.0 - 1 = \) 0.0
NCERT (2026-27 edition):
Abnormal Molar Masses – …
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | – | – | 02-06 S3 31-05 S1 | – |