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The degree of dissociation $(\alpha )$ of a weak electrolyte, $A^{x}B^{y}$ is related to van’t Hoff factor $(i)$ by the expression

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

Option 2)
$\alpha =\frac{i-1}{(x+y+1)}\;$

Option 3)
$\alpha =\frac{(x+y-1)}{i-1}$

Option 4)
$\alpha =\frac{(x+y+1)}{i-1}$

Answers (1)

best_answer

As we learnt in

Vant Hoff factor for dissociation -

$i= 1+(n-1)\alpha$

Where

$n$ is the no. of dissociated particles

$\alpha =$ degree of association

- wherein

$NaC l \: \: \: n = 2$

$CaCl_{2} \: \: \: n = 3$

$K_{4}[F(CN_{6})]\: \: \: \: \: n=5$

$AxBy\rightarrow xA^{y+}+yB^{x-}$

$1-\alpha$ $x\alpha$ $y\alpha$

$i=1+(n-1)\alpha$

Where n = x + y

$i=1+(x+y-1)\alpha$

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

Correct option is 1.

Option 1)

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

This is the correct option.

Option 2)

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

This is an incorrect option.

Option 3)

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

This is an incorrect option.

Option 4)

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

This is an incorrect option.

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