1.25of a mixture of Na_(2)CO_(3) and Na_(2)SO_(4) was dissolvedin 25o mL of water.25 mL of this solution required20 mL of 0.01 " N " H_(2)SO_(4)solution forexact neutralisation .Calculate the percentageof Na_(2)CO_(3) in the mixture .

1.25of a mixture of Na_(2)CO_(3) and Na_(2)SO_(4) was dissolvedin 25o

1.	1.25of a mixture of Na_(2)CO_(3) and Na_(2)SO_(4) was dissolvedin 25o mL of water.25 mL of this solution required20 mL of 0.01 " N " H_(2)SO_(4)solution forexact neutralisation .Calculate the percentageof Na_(2)CO_(3) in the mixture .
Answer» Solution :In this PROBLEM only `Na_(2)CO_(3)`is NEUTRALISED by `H_(2)SO_(4)`. Let the AMOUNT of `na_(2)CO_(3)` be x g ` :. " equivalent of " Na_(2)CO_(3) = x/53 "" …(Eqn.4i)` m.e of `Na_(2)CO_(3) = x/53 XX 1000` ` :." m.e of "Na_(2)CO_(3) ` in 250 mL of the mixture solution = `(1000x)/53 ` ` :. ` m.e of `Na_(2)CO_(3) ` in 25 mL of the mixture solution = `(100X)/53` Now m.e of 25 mL of mixture solution = m.e of 20 mL of 0.1 N `H_(2)SO_(4)`...(Eqn.2) ` (100x)/53 = 0.1 xx 20 ` ` x = 1.06 g ` ` :." % of " Na_(2)CO_(3) =(1.06)/(1.25) xx 100 = 84.8 %`