Question

100ml of solution containing 0.4g of impure sample of $H_2O_2$ completely reacts with 0.632 g of $KMnO_4$ (Acidic). What Should be the percentage purity of $H_2O_2$ ?​

Answer 1

Explanation:

10 ml of 1N KMnO4 solution is used to titrate 0.2gm of H2O2

Mole of H2O2 = 0.2/34= 0.0588 moles

if this solution is made upto 1 ltr to titrate, we have molarity = 0.0588 M

Since H2O2 release 2 H+ ions, Its normality = 2 x 0.0588 = 0.01174 N

Using equation

N1 V1 = N2 V2

10 * 1 = 0.01174 * V2

V2 = 854ml

Hence effective 854 ml is used in 1000 ml depending on purity

hence Purity of H2O2 is 85.4%

Answer 2

$\huge \mathfrak \pink{answer}$

2KMnO4 + H202 = 2KOH + MnO2 +O2

According to the above reaction, 2 moles of

KMnO4 react with H2O2 = 1 mol

158*2g KMnO4 react with H2O2 = 34g

0.632 g KMnO4 react with H2O2 = 34*0.632/(158*2)

= 0.068g

But according to the given ques, mass of H2O2 used for the reaction = 0.4 g

thus, the percentage purity is 0.068*100/0.4 = 17%