Solutions Question 387
Question: The expression relating degree of dissociation of the weak electrolyte $ A_{x}B_{y} $ with its van’t Hoff factor is
Options:
A) $ \alpha =\frac{i-1}{x+y-1} $
B) $ \alpha =\frac{i+1}{x+y-1} $
C) $ \alpha =\frac{x+y-1}{i-1} $
D) $ \alpha =\frac{x+y-1}{i+1} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{1-\alpha }{\mathop{{A_{\times }}B_{y}}},\underset{\alpha x}{\mathop{x{A^{y+}}}},+\underset{\alpha y}{\mathop{y{B^{x-}}}}, $ Total amount of species $ =1+(x+y-1)\alpha $ Van’t Hoff factor, $ i=\frac{1+(x+y-1)\alpha }{1} $ or $ \alpha =\frac{i-1}{x+y-1} $
