`4.35` g of a mixture of NaCl and `Na_(2)CO_(3)` was dissolved in 100 mL of water , 20 mL of which was exactly neutralised by `75.5 ` mL of N/0 solution of `H_(2)SO_(4)` . Calculate the percentage of NaCl and `na_(2)CO_(3)` in the mixture .

Explanation

In this problem ,`Na_(2)CO_(3)` is neutralised by `H_(2)SO_(4)` solution . Let the amount of `Na_(2)CO_(3)` be x g .
` :. ` equivalent of `na_(2)CO_(3)` in 100 mL in solution = `x/53 ` …(Eqn.4i)
` :. ` m.e of `Na_(2)CO_(3)` in 10 mL of solution ` = x/53 xx1000 " " ...(Eqn.3)`
` :. ` m.e of `na_(2)CO_(3)` in 20 mL of solution ` = (1000 x)/53 xx20/100 = (200 x)/53 `
Now , m.e of `H_(2)SO_(4)` solution ` = 1/10 xx 75.50 " " ...(Eqn.1)`
` :. ` m.e of `Na_(2)CO_(3) ` = m.e of `H_(2)SO_(4) " " ...(Eqn.2)`
` :. " " (200x)/53 = 7.55 `
x = 2
` :. " " % of Na_(2)CO_(3) = 2/(4.35) xx 100 = 45.99 % `
` % of naCl = 100 - 45.99 = 54.01 % `