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100 mL of mixture of NaOH and Na(2)SO(4)...

`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

A

0.2 g

B

0.4 g

C

0.6 g

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
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`
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