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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» <html><body><p><br/></p>Solution :Suppose, the <a href="https://interviewquestions.tuteehub.com/tag/amount-374803" style="font-weight:bold;" target="_blank" title="Click to know more about AMOUNT">AMOUNT</a> of NaCI in the mixture is <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> g. .-. The amount of <a href="https://interviewquestions.tuteehub.com/tag/kci-527832" style="font-weight:bold;" target="_blank" title="Click to know more about KCI">KCI</a> in the mixture = 0.9031 - x g The corresponding <a href="https://interviewquestions.tuteehub.com/tag/chemical-914796" style="font-weight:bold;" target="_blank" title="Click to know more about CHEMICAL">CHEMICAL</a> equations are<br/> `underset(2 xx 58.44 g)(2NaCl) + H_(2)SO_(4) to underset(142.04 g) + 2HCl` <br/> `underset(2 xx 74.55 g)(2KCl) + H_(2)SO_(4) to underset(174.26 g)(K_(2)SO_(4)) + 2HCl` <br/> `therefore 2 xx 58.44` of NaCl give `Na_(2)SO_(4) = 142.04 g` <br/> `therefore x g` of NaCl will give `Na_(2)SO_(4) =(142.04)/(2 xx 58.44) xx x` <br/> 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)` <br/> `therefore` The amount of the resulting mixture of sulphates =1.0784 g <br/> `therefore 142.04/(2 xx 58.44 xx x) + [174.26/(2 xx 74.55) xx (0.9031 -x)] = 1.0784` <br/> which gives x=0.4829 g <br/> `therefore` Percentage of NaCl in the given mixture <br/> `=0.4829/0.9031 xx 100 = 53.47` <br/> and the percentage of KCI in the mixture `=100-53.47 = 46.53`</body></html>


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