The percent strength (W/V) and normality of 15 mL of a 40 volume${\mathrm{H}}_2{\mathrm{O}}_2$ solution is respectively.

 

. 40 volume ${\mathrm{H}}_2{\mathrm{O}}_2$means that : 

1 mL of 40 volume ${\mathrm{H}}_2{\mathrm{O}}_2$ produces ${\mathrm{O}}_2=40 \mathrm{mL}$

$\therefore 15\mathrm{mL}\text{ of }40\text{ volume }{\mathrm{H}}_2{\mathrm{O}}_2\text{ produces }$

                           ${\mathrm{O}}_2=\frac{40\mathrm{mL}}{1\mathrm{mL}}\times 15\mathrm{mL}=600\mathrm{mL}$

Reaction : $\underset{\begin{array}{l}2\left[\right(2\times 1)+(2\times 16\left)\right]\\ =68\mathrm{g}\end{array}}{2{\mathrm{H}}_2{\mathrm{O}}_2}\to 2{\mathrm{H}}_2\mathrm{O}+\underset{22400\mathrm{mLat}\text{ N.T.P. }}{ }$

22400 mL of ${\mathrm{O}}_2$ at N.T.P. is obtained from 

                         ${\mathrm{H}}_2{\mathrm{O}}_2=68\mathrm{g}$

$\therefore 600\mathrm{mL}\text{ of }{\mathrm{O}}_2\text{ at N.T.P. }$is obtained from

${\mathrm{H}}_2{\mathrm{O}}_2=\frac{68\mathrm{g}}{22400\mathrm{mL}}\times 600\mathrm{mL}=1.82\mathrm{g}$${\mathrm{H}}_2{\mathrm{O}}_2=\frac{1.82\mathrm{g}}{15\mathrm{mL}}\times 100\mathrm{mL}$

(i) 15 mL of given ${\mathrm{H}}_2{\mathrm{O}}_2$ solution contains 

                    ${\mathrm{H}}_2{\mathrm{O}}_2=1.82\mathrm{g}$

$1000\mathrm{mL}\text{ (or }1\mathrm{L}\text{ ) of given }{\mathrm{H}}_2{\mathrm{O}}_2\text{ solution contains }{\mathrm{H}}_2{\mathrm{O}}_2$

$=\frac{1.82\mathrm{g}}{15\mathrm{mL}}\times 1000\mathrm{mL}=121.3{\mathrm{gL}}^{-}$

$\therefore$ Strength of given 40 volume

${\mathrm{H}}_2{\mathrm{O}}_2\text{ solution }=121.3{\mathrm{gL}}^{-}$

Similarly : 15 mL of given ${\mathrm{H}}_2{\mathrm{O}}_2$ solution contains

                 ${\mathrm{H}}_2{\mathrm{O}}_2=1.82\mathrm{g}$

$\therefore$ 100 mL of given  ${\mathrm{H}}_2{\mathrm{O}}_2$solution contain.

           ${\mathrm{H}}_2{\mathrm{O}}_2=\frac{1.82\mathrm{g}}{15\mathrm{mL}}\times 100\mathrm{mL}$

$=12.13\mathrm{g}\text{ per }100\mathrm{mL}$

$\therefore$Percent strength of given

${\mathrm{H}}_2{\mathrm{O}}_2=12.13\mathrm{\%}(\mathrm{W}/\mathrm{V})$

$\text{ (ii) Eq. wt. of }\begin{array}{r}{\mathrm{H}}_2{\mathrm{O}}_2=\frac{\text{ Mol. wt. of }{\mathrm{H}}_2{\mathrm{O}}_2}{2}\\ =\frac{(2\times 1)+(2\times 16)}{2}=17\end{array}$

                                           $\left[\because {\mathrm{H}}_2{\mathrm{O}}_2\to 2{\mathrm{H}}^{+}+{\mathrm{O}}_2+2e^{-}\right]$

$\begin{array}{c}\therefore \text{ Normality }=\frac{{\text{ Strength in gL }}^{-}}{\text{Eq.wt. }}=\frac{12.13}{17}\\ =7.14\mathrm{N}\end{array}$

.So, the correct answer is (3).