Equilibrium Question 354
Question: Degree of dissociation of $ 0.1,NCH_3COOH $ is (Dissociation constant $ =1\times {10^{-5}} $ ) [MP PET 1997]
Options:
A) $ {10^{-5}} $
B) $ {10^{-4}} $
C) $ {10^{-3}} $
D) $ {10^{-2}} $
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Answer:
Correct Answer: D
Solution:
Degree of dissociation a = - Normality of solution = 0.1 N $ =\frac{1}{10}N $ Volume = 10 litre Dissociation constant $ K=1\times {10^{-5}} $
$ K=\frac{{{\alpha }^{2}}}{V} $ ; $ \alpha =\sqrt{KV} $
$ =\sqrt{1\times {10^{-5}}\times 10} $ ; $ \alpha =1\times {10^{-2}} $
