0.9031 g of a mixture of NaCI and KCI on treatment with H_2SO_4 yielded 1.0784 g of a mixture of Na_2SO_4 and K_2SO_4.Calculate the percent composition of the mixture.

0.9031 g of a mixture of NaCI and KCI on treatment with H_2SO_4 yielde

1.	0.9031 g of a mixture of NaCI and KCI on treatment with H_2SO_4 yielded 1.0784 g of a mixture of Na_2SO_4 and K_2SO_4.Calculate the percent composition of the mixture.
Answer» Solution :Suppose, the AMOUNT of NaCI in the mixture is X g. .-. The amount of KCI in the mixture = 0.9031 - x g The corresponding CHEMICAL equations are `underset(2 xx 58.44 g)(2NaCl) + H_(2)SO_(4) to underset(142.04 g) + 2HCl` `underset(2 xx 74.55 g)(2KCl) + H_(2)SO_(4) to underset(174.26 g)(K_(2)SO_(4)) + 2HCl` `therefore 2 xx 58.44` of NaCl give `Na_(2)SO_(4) = 142.04 g` `therefore x g` of NaCl will give `Na_(2)SO_(4) =(142.04)/(2 xx 58.44) xx x` Similarly, the amount of `K_2So_4` formed by (0.9031 x) g of KCl `=(174.26)/(2 xx 74.55) xx (0.9031 -x)` `therefore` The amount of the resulting mixture of sulphates =1.0784 g `therefore 142.04/(2 xx 58.44 xx x) + [174.26/(2 xx 74.55) xx (0.9031 -x)] = 1.0784` which gives x=0.4829 g `therefore` Percentage of NaCl in the given mixture `=0.4829/0.9031 xx 100 = 53.47` and the percentage of KCI in the mixture `=100-53.47 = 46.53`