Volume of ${\mathrm{H}}_2{\mathrm{O}}_2=5 \mathrm{cc}$

Weight of ${\mathrm{I}}_2$ liberated = $0.508 \mathrm{g}$

Reaction of ${\mathrm{H}}_2{\mathrm{O}}_2$ with $\mathrm{KI}$,

${\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{I}}^{-}\to {\mathrm{I}}_2+{\mathrm{H}}_2\mathrm{O}$.

Here, valency factor of iodine $\left(2{\mathrm{I}}^{-}\to {\mathrm{I}}_2\right)=$Change in oxidation number per mole of iodine = $2$.

Equivalent weight of ${\mathrm{I}}_2=\frac{\mathrm{Molecular} \mathrm{weight} \mathrm{of} {\mathrm{I}}_2}{\mathrm{valency} \mathrm{factor}}=\frac{254}{2}=127 \mathrm{g}$.

$\mathrm{Milli} \mathrm{equivalents}=$ $\frac{\mathrm{Given} \mathrm{weight} \mathrm{of} {\mathrm{I}}_2}{\mathrm{Equivalent} \mathrm{weight} \mathrm{of} {\mathrm{I}}_2}=\frac{0.508}{127}$

$=$Normality$\times$Volume.

Now, 

$\mathrm{Milli} \mathrm{equivalents}$ of ${\mathrm{H}}_2{\mathrm{O}}_2=$$\mathrm{Milli} \mathrm{equivalents}$ of ${\mathrm{I}}_2$

$\underset{\left({\mathrm{H}}_2{\mathrm{O}}_2\right)}{{\mathrm{N}}_1{\mathrm{V}}_1}=\underset{\left({\mathrm{I}}_2\right)}{{\mathrm{N}}_2{\mathrm{V}}_2}$

Substituting the values, we get:

$$\begin{gathered} {\mathrm{N}}_1\times \frac{5}{1000}=\frac{0.508}{127} \\ \Rightarrow {\mathrm{N}}_1=\frac{0.508\times 1000}{5\times 127}=0.8 \mathrm{N}. \end{gathered}$$

Now,

for ${\mathrm{H}}_2{\mathrm{O}}_2$,

$5.6\times$ Normality $=$Volume strength

$\Rightarrow 5.6\times 0.8=4.48 \mathrm{V}$.