1.94 g of a mixture of KOH (56) and K_(2)CO_(3)(138) is dissolved in water and separated into two equal parts by volume. One part required 50 mL 0.1 M H_(2)SO_(4) to reach the phenolphthalein end point while the other part required 75 mL of the same acid to reach the methyl orange end point. The mass percentage of K_(2)CO_(3) in the mixture is

1.94 g of a mixture of KOH (56) and K_(2)CO_(3)(138) is dissolved i

1.	1.94 g of a mixture of KOH (56) and K_(2)CO_(3)(138) is dissolved in water and separated into two equal parts by volume. One part required 50 mL 0.1 M H_(2)SO_(4) to reach the phenolphthalein end point while the other part required 75 mL of the same acid to reach the methyl orange end point. The mass percentage of K_(2)CO_(3) in the mixture is
Answer» `35.5%` `71%` `29%` `64.5%` Solution :Let one PART of solution has x mmol KOH and y mmol `K_(2)CO_(3)`. `implies x + y = 10` and x + 2y = 15 `implies y = 5, x = 5` Total mmol of `K_(2)CO_(3)` in the sample = 10 MASS of `K_(2)CO_(3)=(10 XX 138)/(1000)= 1.38 g` `implies` Mass % of `K_(2)CO_(3)= (1.38)/(1.94) xx 100 = 71%`