Excess of Kl and dil. H_2SO_4 were mixed in 50 mL H_2O_2. The liberated l_2 required 20 mL 0.1 N Na_2S_2O_3. Find out the strength of H_2O_2 in g "litre"^(-1).

Excess of Kl and dil. H_2SO_4 were mixed in 50 mL H_2O_2. The liberate

1.	Excess of Kl and dil. H_2SO_4 were mixed in 50 mL H_2O_2. The liberated l_2 required 20 mL 0.1 N Na_2S_2O_3. Find out the strength of H_2O_2 in g "litre"^(-1).
Answer» SOLUTION :The corresponding chemical equations are: `2Kl + underset("one mole" 34.016 g)(H_(2)SO_(4)) + H_(2)O_(2) to K_(2)SO_(4) + 2H_(2)O + underset("one mole")(I_(2))` `underset("two moles" 2 xx 158.1 g)(2Na_(2)S_(2)O_(3)) + underset("one mole" 253.8 g)(I_(2)) to Na_(2)S_(4)O_(6) + 2NaI` WEIGHT of `Na_2S_2O_3` present in 20 mL 0.1 N solution can be obtained as follows : `therefore w =(NEV)/1000` `therefore w =(0.1 xx 158.1 xx 20)/1000 = 0.3162 g` `therefore` Amount of `H_(2)O_(2)` in one LITRE `=(0.03401)/50 xx 1000 = 0.6802g` Hence, the STRENGTH of given `H_(2)O_(2)` solution is `0.6802 g L^(-1)`