Question

what is the weight of available oxygen from a solution of H2 O2 if 20 ml of the solution needs 25 ml N/20 KMnO4 for complete oxidation​

Answer 1

The available weight of oxygen as 2g per liter.

Explanation:

The oxidation reaction is as

$2KMnO_{4}+5H_{2}O_{2}+4H_{2}SO_{4}$→$2KHSO_{4}+8H_{2}O+5O_{2}$

$Meq. of O_{2} =Meq. of H_{2} O_{2} =Meq.of KM_{n} O_{4}$

It is given that $10 ml$ of $H_{2} O_{2}$ solution when treated with $25ml$ of $\frac{N}{20} KM_{n} O_{4}$

for complete oxidation.

$N*20=25*\frac{1}{20}$

Normality of $H_{2} O_{2} =0.0625 N=Normality of O_{2}$

Weight of $O_{2}$$=0.0625*32$

$=2g per liter$

Therefore, the available weight of oxygen is 2g per liter.