The degree of dissociation ( $\alpha$ ) of a weak electrolyte, $A_{x} B_{y}$ is related to van't Hoff factor (i) by the expression :

(D) $\alpha=\frac{i-1}{(x+y-1)}$

JEE Main · 2011 · 10 Apr 2011 · Offline · Solutions

The degree of dissociation (

α

) of a weak electrolyte,

A_{x} B_{y}

is related to van't Hoff factor (i) by the expression :

(A) $α = \frac{i - 1}{x + y + 1}$
(B) $α = \frac{x + y - 1}{i - 1}$
(C) $α = \frac{x + y + 1}{i - 1}$
(D) $α = \frac{i - 1}{( x + y - 1 )}$ ✓ Correct

Answer: (D)

Solution

For the weak electrolyte

A_{x} B_{y}

dissociating into

x + y

ions, the van't Hoff factor is

i = 1 + α (n - 1)

with

n = x + y

. Rearranging,

i - 1 = α (x + y - 1)

, so the degree of dissociation is

α = \frac{i - 1}{x + y - 1}

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option

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