10.875 g of a mixture of NaCI and Na_2CO_3 was dissolved in water and the volume was made up to 250 mL. 20.0 mL of this solution required 75.5 mL of N/10 H_2SO_4. Find out the percentage composition of the mixture.

10.875 g of a mixture of NaCI and Na_2CO_3 was dissolved in water and

1.	10.875 g of a mixture of NaCI and Na_2CO_3 was dissolved in water and the volume was made up to 250 mL. 20.0 mL of this solution required 75.5 mL of N/10 H_2SO_4. Find out the percentage composition of the mixture.
Answer» Solution :NaCl does not react with `H_2SO_4`. Therefore, whole of the sulphuric acid reacted with `Na_2CO_3` only. Suppose, 10.875 g of the mixture contain x g of `Na_2CO_3`. This is present in 250 ML of the solution. Since, `w =(NEV)/1000` We have `x=(N xx 53 xx 250)/1000` or `N =(1000 x)/(53 xx 250) = (1000 x)/(13250)` According to the normality relation, `underset(Na_(2)CO_(3) "sol")(N_(1)V_(1)) = underset(H_(2)SO_(4))(N_(2)V_(2))` `=(1000 x)/(13250) xx 20.0 =1/10 xx 75.5` or `x=(13250 xx 75.5)/(1000 xx 20.0 xx 10) = 5.0018 g` Thus, 10.875 g of the sample contain 5.0018 g of `Na_(2)CO_(3)` `therefore` MASS of NaCl present `=10.875 - 5.0018 = 5.8732 g` Hence, the amount of `Na_2CO_3`in the sample `=5.0018/10.875 xx 100 = 45.994 %` and the amount of NaCl in the sample `=5.6732/10.875 xx 100 = 54.006%`