`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,the amount of 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

Moles of `H_(2)SO_(4)` used `= (10)/(1000) xx 0.5 = 0.005 rArr` Moles of `H^(+)` ions used `= 2 xx 0.005 = 0.01`
`rArr` Moles of NaOH used = 0.01 (As `Na_(2)SO_(4)` is neutral being a salt of strong acid and strong base)
Mass of NaOH used `= 0.01 xx 40 = 0.4 g`