Solutions Question 265
Question: The degree of dissociation (1) of a weak electrolyte $ A_{x}B_{y} $ is related to van’t Hoff factor (i) by the expression
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:
van’t Hoff factor (i) is related degree of dissociation ( $ \alpha $ ) as $ \alpha =\frac{i-1}{n-1} $ Here, n are the moles of an electrolyte $ ( A-B ) $ dissolve in a solvent. For $ A_{x}B_{y},A_{x}B_{y}\rightarrow x{A^{+y}}+y{B^{-x}} $ Therefore, $ \alpha =\frac{i-1}{(x+y-1)} $
