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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» <html><body><p></p>Solution :NaCl does not react with `H_2SO_4`. Therefore, whole of the sulphuric acid reacted with `Na_2CO_3` only. <br/> Suppose, <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>.875 g of the mixture contain x g of `Na_2CO_3`. This is present in 250 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> of the solution.<br/> Since, `w =(NEV)/1000` <br/> We have `x=(N xx 53 xx 250)/1000` <br/> or `N =(1000 x)/(53 xx 250) = (1000 x)/(13250)` <br/> According to the normality relation, <br/> `underset(Na_(2)CO_(3) "sol")(N_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)V_(1)) = underset(H_(2)SO_(4))(N_(2)V_(2))` <br/> `=(1000 x)/(13250) xx 20.0 =1/10 xx 75.5` <br/> or `x=(13250 xx 75.5)/(1000 xx 20.0 xx 10) = 5.0018 g` <br/> Thus, 10.875 g of the sample contain 5.0018 g of `Na_(2)CO_(3)` <br/> `therefore` <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of NaCl present `=10.875 - 5.0018 = 5.8732 g` <br/> Hence, the amount of `Na_2CO_3`in the sample `=5.0018/10.875 xx 100 = 45.994 %` <br/> and the amount of NaCl in the sample `=5.6732/10.875 xx 100 = 54.006%`</body></html>


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