`100 mL` of mixture of `NaOH` and `Na_(2)SO_(4)` is neutralised by `10 mL` of `0.5 M H_(2) SO_(4)`. Hence, NaOH in `100 mL` solution is

Options

A. `0.2 g`
B. `0.4 g`
C. `0.6 g`
D. none of these

Answer

B

Explanation

As `Na_(2)SO_(4)` is already neutral so only `NaOH` will be neutralized by `H_(2)SO_(4)`
`:. Eq. of NaOH=Eq. of H_(2)SO_(4)`
`(wxx1000)/(40xx100)xx1xx(100)/(1000)=10xx10^(-3)xx0.5xx2`
`W=40xx10^(-2)=0.4 g`