Two litre of `NH_(3)` at `30^(@)C` and 0.20 atmosphere is neutralised by `134 mL` of a solution of `H_(2) SO_(4)`. Calculate the normality of `H_(2) SO_(4)`.

Answer

A, B

Explanation

`PV = (W)/(Mw) RT`
Mole `= (W)/(Mw) = (PV)/(RT) = (0.2 xx 2)/(0.0821 xx 303) = 0.01608 mol`
= 16.08 mmol or mEq
`({:(1 "mole of" NH_(3) = 1 g Eq. of NH_(3)),(1 "mmole of" NH_(3) = 1 mEq of NH_(3)):})`
`N xx V = mEq`
`N xx 134 mL = 16.08`
`:. N = 0.12`