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)

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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