Question

25 mL of hydrogen peroxide solution was added to excess of acidified potassium iodide solution. The iodine so liberated required 20 mL of 0.1 N sodium thiosulphate solution , calculate the strength interns of normality, percentage and volume.

Answer 1

Let us assume the normality of Hydrogen peroxide be N.

Now,A/Q
20 ml of 0.1 N Na2S2O3 = 20 ml of I2 = 25 ml of N H2O2

Taking the relationship of only iodine and hydrogen peroxide,

20 ml of I2 = 25 ml of N H2O2
20 x 0.1 = 25 N
N = 2÷25 = 0.08

So strength in terms of normality = Normality x Equivalent Mass = 0.08 x 17 = 1.36 g/L

In terms of percent% = 1.36/1000 x 100 = 0.136

From this we infer that,
For 100 ml of solution 0.136 g of H2O2 is present.
For 1 ml 0.136÷100 = 0.00136 g of H2O2 is present.

For 68 g of H2O2, volume of oxygen given = 22.4 L or 22400 ml (NTP)

So,
0.00136 g of H2O2 , volume of oxygen = 22400÷68 x 0.00136 = 0.448 ml

Strength in terms of normality = 0.08 N
Percentage = 0.136%
Volume = 0.448 ml

Hope This Helps You!

Answer 2

M of H2O2 = M of Na2S2O3

N=Normality of H2O2

N×25= 0.3×20N = 0.24

Volume = N×5.6=0.24×5.6= 1.344g/L