Q27 / 150 · MHT CET 2026 (19 April Shift 1) · Chemistry
van't Hoff factor for $\text{BaCl}_2$ is $2.47$, calculate the percentage dissociation of in its aqueous solution.
- A$25.6\%$
- B$35.7\%$
- Correct answerC$73.5\%$
- D$13.6\%$
Solution
The dissociation reaction of $\text{BaCl}_2$ is:
$\text{BaCl}_2 \rightleftharpoons \text{Ba}^{2+} + 2\text{Cl}^{-}$
The number of ions produced per formula unit, $n = 3$.
The relation between van't Hoff factor $i$ and degree of dissociation $\alpha$ is:
$i = 1 + (n - 1)\alpha$
Substituting the given values:
$2.47 = 1 + (3 - 1)\alpha$
$2.47 = 1 + 2\alpha$
$2\alpha = 1.47$
$\alpha = 0.735$
Percentage dissociation $= \alpha \times 100 = 73.5\%$
$\text{BaCl}_2 \rightleftharpoons \text{Ba}^{2+} + 2\text{Cl}^{-}$
The number of ions produced per formula unit, $n = 3$.
The relation between van't Hoff factor $i$ and degree of dissociation $\alpha$ is:
$i = 1 + (n - 1)\alpha$
Substituting the given values:
$2.47 = 1 + (3 - 1)\alpha$
$2.47 = 1 + 2\alpha$
$2\alpha = 1.47$
$\alpha = 0.735$
Percentage dissociation $= \alpha \times 100 = 73.5\%$
Hence, the correct answer is option: 'C'.