Solutions Question 265

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} $

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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)} $



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