Question

Vant hoff factor of $BaCl_{2}$ of conc. $0.01M$ is 1.98. Percentage dissociation of $BaCl_{2}$ on this conc. Will be.

• A. 49
• B. 69
• C. 89
• D. 98 (e) 100

Answer 1

Answer: (A)

49

Answer 2

$BaCl_{2}$ ⇌ $Ba^{2 +}$ + $2Cl^{-}$

Initially 1 0 0

After dissociation $a - \alpha$ $\alpha$ $2\alpha$

Total = $1 - \alpha + \alpha + 2\alpha = 1 + 2\alpha$

$\alpha = \frac{1.98 - 1}{\alpha} = \frac{0.98}{\alpha} = 0.49$

for a mole $\alpha = 0.49$

For 0.01 mole $\alpha = \frac{0.49}{0.01} = 49$.