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

Options

A. `0.12`
B. `0.23`
C. `0.17`
D. `0.19`

Answer

A

Explanation

For `N H _(3) , PV = (w)/(m) RT`
`therefore (w)/(m)= (PV)/(RT) = (0.2 xx 2)/( 0.0821 xx 303) = 0.01608 therefore `Mole of `NH _(3)=` Equivalent of `NH _(3) =0.01608`
`therefore" "` Meq. Of `NH _(3) = 16. 08` Now Meq. Of `H _(2) S _(4) =` Meq. of `NH _(3)`
`therefore N xx 134 = 16. 08 " " N = 0.12`