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

A sample containing 0.496 gm of (NH_(4))_(2) C_(2)O_(4) (MW = 124) and

1.	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
Answer» `40%` `50%` `60%` `75%` Solution :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%`