Volume strength of H2O2 using iodometric titration

[Titan97]

Homework Statement

To a 25ml H2O2 solution, excess acidified solution of potassium iodide is added. The libereated iodine required 20ml of 0.3N sodium thiosulphate solution. Calculate the volume strength of H2O2.

Homework Equations

Normality = n-factor × Molarity
Volume strength = Molarity × 11.2

The Attempt at a Solution

I will use moles and not equivalence. But to do that, I need to convert normality to molarity.

For the reaction of sodium thiosulphate with iodine, n-factor is 2 for sodium thiosulphate.
Hence M = N/n = 0.15.
So number of milli moles of thiosulphate is 3.

$$I_2 + 2Na_2S_2O_3 \rightarrow 2Na_2S_4O_6 + 2NaI$$

So number of millimoles of iodine is $\frac{3}{2}$.

$$2H^+ + H_2O_2 + 2KI \rightarrow 2H_2O+I_2+2K^+$$

Hence millimoles of hydrogen peroxide is $\frac{3}{2}$

So molarity of hydrogen peroxide is $\frac{3}{50}$

So volume strength = $\text{M}\times 11.2 = 0.672$

But the answer given is double of what I got.

 

[Titan97]

oops...
The number of electrons exchanged by 1 mole of thiosulphate is 1.
$$2e^-+2S_2O_3^{2-}\rightarrow S_4O_6^{2-}$$

Hence n-factor is 1 and not 2. Now I got the given answer.