When `0.004 M Na_(2)SO_(4)` is an isotonic acid with `0.01 M `glucose, the degree of dissociation of `Na_(2)SO_(4)` is

Options

A. `75%`
B. `50%`
C. `25%`
D. `85%`

Answer

A

Explanation

For `Na_(2)SO_(4)`,`pi_(1)=iMRT`
`=ixx 0.004 xx RT`
For glucose,`pi_(2)=CRT`
`=0.01 RT`
`pi_(1) =pi_(2)`(isotonic)
or `0.004i =0.01`
`i=2.5`
Total particles =`1-alpha + 2alpha +alpha`
`1+2alpha`
`:.i=(1 + 2alpha)/1=2.5`
`alpha=0.75` or `75%` dissociated.