Equilibrium Question 198
Question: Vant hoff factor of $ BaCl_2 $ of conc. $ 0.01M $ is 1.98. Percentage dissociation of $ BaCl_2 $ on this conc. Will be [Kerala CET 2005]
Options:
49
69
89
98
100
Show Answer
Answer:
Correct Answer: A
Solution:
$ BaCl_2 $ ⇌ $ B{a^{2+}} $ + $ 2C{l^{-}} $
| 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 $
BETA
