Strength of a ${\mathrm{H}}_2{\mathrm{O}}_2$ solution is labeled as $1.79 \mathrm{N}$. Its strength can also be expressed as closest to

$2{\mathrm{H}}_2{\mathrm{O}}_{2\left(\mathrm{aq}.\right)}⟶2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{\ell }\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$

General Solution :- 

It is a common trend that $\mathrm{n}$-factor $\left({\mathrm{H}}_2{\mathrm{O}}_2\right)$ is
taken as '$2$' 

By the definition of volume strength of ${\mathrm{H}}_2{\mathrm{O}}_2$ if we consume $1\ell$ of $1 \mathrm{N} {\mathrm{H}}_2{\mathrm{O}}_2$ in the above equation we are using 1 gram equivalent of
${\mathrm{H}}_2{\mathrm{O}}_2\equiv 0.5$ moles of ${\mathrm{H}}_2{\mathrm{O}}_2$ (by using $\mathrm{n}$-factor $=2$)

This will produce $\frac{1}{4}$ moles of ${\mathrm{O}}_2$ gas at $\mathrm{N}.\mathrm{T}.\mathrm{P}.$

$\equiv \frac{1}{4}\times 22.4=5.6\ell$ of ${\mathrm{O}}_2$ gas

i.e. $1\ell ,1 \mathrm{N} {\mathrm{H}}_2{\mathrm{O}}_2$ solution gives $5.6\ell$ ${\mathrm{O}}_2$ at $\mathrm{N}.\mathrm{T}.\mathrm{P}$

Hence $1 \mathrm{N}\equiv 5.6$ 'vol.' ${\mathrm{H}}_2{\mathrm{O}}_2$ solution

In the given question it is $1.76 \mathrm{N} {\mathrm{H}}_2{\mathrm{O}}_2$ solution Hence volume strength $=5.6\times 1.79\simeq 10$ volumes

Suggested Solution :- 

$2{\mathrm{H}}_2{\mathrm{O}}_{2\left(\mathrm{aq}.\right)}⟶2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{\ell }\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$

The above equation is a classic example of
disproportionation of ${\mathrm{H}}_2{\mathrm{O}}_2$, hence we should take $\mathrm{n}$-factor of ${\mathrm{H}}_2{\mathrm{O}}_2=1$

$1\mathrm{\ell },1 \mathrm{N} {\mathrm{H}}_2{\mathrm{O}}_2$ solution $\equiv 1 \mathrm{Gm}$ equivalent of ${\mathrm{H}}_2{\mathrm{O}}_2$

$\equiv 1$ mole of ${\mathrm{H}}_2{\mathrm{O}}_2$

This will produce $\frac{1}{2}$ moles of ${\mathrm{O}}_2$ gas at $\mathrm{N}.\mathrm{T}.\mathrm{P}.$

$=\frac{1}{2}\times 22.4=11.2\ell$ of ${\mathrm{O}}_2$ gas

i.e. $1\ell ,1 \mathrm{N} {\mathrm{H}}_2{\mathrm{O}}_2$ solution gives $11.2\ell$ of ${\mathrm{O}}_2$ gas at $\mathrm{N}.\mathrm{T}.\mathrm{P}.$

Hence $1 \mathrm{N}\equiv 11.2$ 'vol' ${\mathrm{H}}_2{\mathrm{O}}_2$ solution 

In the given question it is $1.79 {\mathrm{NH}}_2{\mathrm{O}}_2$ solution

Hence, volume strength $=1.79\times 11.2\simeq 10$ volumes

Suggested Answer is option (A)