A `0.004M` solution of `Na_(2)SO_(4)` is isotonic with a `0.010 M` solution of glucose at same temperature. The apparent degree of dissociation of `Na_(2)SO_(4)` is

Options

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

Answer

C

Explanation

`pi(Na_(2)SO_(4))=iCRT=i(0.004)RT`
`pi"(Glucose) = CRT = 0.010 RT"`
As solution are isotonic, I (0.004) RT = 0.01 RT.
This gives I = 2.5
`{:("Now ",Na_(2)SO_(4),hArr,2Na^(+),+,SO_(4)^(2-)),(,"1 mole",,0,,0),(,1-alpha,,2alpha,,alpha","):}`
`" Total "=1+2alpha`
`therefore" "i=1+2alpha`
`"or "alpha=(i-1)/(2)=(2.5-1)/(2)=0.75=75%`.