`1.25` of a mixture of `Na_(2)CO_(3) and Na_(2)SO_(4)` was dissolved in 25o mL of water. 25 mL of this solution required 20 mL of `0.01 " N " H_(2)SO_(4) ` solution for exact neutralisation .Calculate the percentage of `Na_(2)CO_(3)` in the mixture .

Explanation

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 %`