Equilibrium Question 348

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:

A) 49

B) 69

C) 89

D) 98

E) 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 $



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