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1.2 g mixture of Na_(2)CO_(3) and K_(2)CO_(3) was dissolved in water to form 100 cm^(3) of the solution. 20 cm^(3) of this solution required 40 cm6(3) of 0.1 N HCl solution for neutralisation. Calculate the percentage composition of the mixture. |
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Answer» Solution :Step I. Calculation of TOTAL no of gram equivalents in the mixture Let the mass of `Na_(2)CO_(3)` in the mixture = x g `:.` Mass of `K_(2)CO_(3)` in the mixture `= (1.2 - x)g` Equivalent mass of `Na_(2)CO_(3)=("MOLECULAR mass")/(2)=(106)/(2)=53 ` Equivalent mass of `K_(2)CO_(3)=("Molecular mass")/(2)=(138)/(3)=69` No. of gram equivalents of `Na_(2)CO_(3)=(x)/(53)` No. of gram equivalent of `K_(2)CO_(3)=((1.2 - x))/(69)` Total no. of gram equivalent of `Na_(2)CO_(3)` and `K_(2)CO_(3) = (x)/(53) +((1.2 -x))/(69)` Step II. Calculation of the no. of gram equivalents by titration No. of gram equivalents in `40 cm^(3)` of 0.1 HCl `= (0.1)/(1000) xx 40 = 4 xx 10^(-3)` `20 cm^(3)` of the mixture solution NEUTRALISE `HCl = 4 xx 10^(-3) g eq`. `100 cm^(3)` of the mixture solution neutralise `HCl = (4 xx 10^(-3))/(20) xx 100 = 2 xx 10^(-2) g aq.` Since acids and bases react in proportions of their gram equivalents `:.` No. of gram equivalents of `Na_(2)CO_(3) and K_(2)CO_(3)=2 xx 10^(-2)` Step III. Calculation of percentage composition of the mixture Equating eqns. (i) and (ii), `(x)/(53)+((1.2-x))/(69)=2xx10^(-2)=0.02` `69x+63.6-53x=0.02xx53xx69=73.14` `16x=73.14-63.6=98.54` `:. x =(9.54)/(16)=0.596g` % of `Na_(2)CO_(3)` in the mixture `= (0.596)/(1.2) xx 100 = 49.67` % of `K_(2)CO_(3)` in the mixture `= 100-49.67 = 50.33`. |
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