A sample containing `0.496 gm` of `(NH_(4))_(2) C_(2)O_(4) (MW = 124)` and inert material was dissolved in water and made strongly alkaline with KOH which converts `NH_(4)^(+)` into `NH_(3)`. The liberated `NH_(3)` was distilled into exactly 50ml of `0.05M H_(2)SO_(4)`. The excess `H_(2)SO_(4)` was back titrated with 10ml of `0.1MNaOH`. The percentage of `(NH_(4))_(2) C_(2)O_(4)` with sample is

Options

A. `40%`
B. `50%`
C. `60%`
D. `75%`

Answer

B

Explanation

m. eqts of excess `H_(2)SO_(4) = 10 xx 0.1 = 1`
m.eqts of `H_(2)SO_(4)` taken `= 50 xx 0.05 xx 2 = 5`
m.eqts of `H_(2)SO_(4)` reacted with
`NH_(4) = 4=` m.eqts of `(NH_(4))_(2) C_(2)O_(4)`
wt of `(NH_(4))_(2)C_(2)O_(4) = 4 xx 62 xx 10^(-3)`
% purity `=(4 xx 0.062)/(0.496) xx 100 = 50%`