Stoichiometry and iodometric titrations $1.25~\mathrm g$ of a sample of bleaching powder is dissolved in $100\ \mathrm{mL}$ of water. $25\ \mathrm{mL}$ of it are treated with $\ce{KI}$ solution. The iodine so liberated required $12.5~ \mathrm{mL}$ of $\frac{\mathrm{M}}{25}$ hypo solution ($\ce{Na2S2O3}$) for complete titration. Find percentage of available chlorine from the sample of bleaching powder.


My attempt:

In $25~\mathrm{mL}$ sample:

mEq of $\ce{Cl2}$ in bleaching powder $=$ mEq of $\ce{I2}$ liberated $=$ mEq of hypo solution

$\mathrm{mEq}$ of Hypo $= NV = \left(1\times\frac{1}{25}\right)\times12.5 = 0.5$

This is where I have a problem, if we are to go by the solution provided in my book, then the n-factor for the hypo solution is 1, whereas, when I write the balanced redox reaction, that is,

$\ce{2S2O3^2- -&gt; S4O6^2- +2e^-}$

You can see that n-factor should be $2$.
No. The $n-$factor is still $1$ because one mole of $\ce{S2O3^2-}$ furnishes 1 mole of electrons. 

In your balanced reaction, 2 moles of $\ce{S2O3^2-}$ gives 2 moles of electrons. To find n-factor, you need to calculate the amount of electrones given by one mole of $\ce{S2O3^2-}$.