i.  $\begin{array}{r}\mathrm{mEq}\text{ of }{\mathrm{H}}_2{\mathrm{O}}_2=\mathrm{mEq}\text{ of }{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\\ (\mathrm{n}=2) \end{array}\begin{array}{r}(\mathrm{n}=2)\\ \left(x\right)\end{array}$

$20\mathrm{mL}\times {\mathrm{N}}_1=40\mathrm{mL}\times {\mathrm{N}}_2$

a. $\mathrm{mEq}\text{ of }{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\equiv \mathrm{mEq}\text{ of }{C}_2{\mathrm{O}}_4^{2-}$   . . . . . (i)

                                             (n=2)

$2.0\times {\mathrm{N}}_2\equiv 5.0\times 1.0\times 2$

${\mathrm{N}}_2\left({\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\right)=5$                             . . . .(ii)

Substituting the value of N₂ from (ii) in (i)

$\begin{array}{r}20\mathrm{mL}\times {\mathrm{N}}_1=40\mathrm{mL}\times 5\\ {\mathrm{M}}_1\left({\mathrm{H}}_2{\mathrm{O}}_2\right)=\frac{10}{2}=5\mathrm{M}\end{array}$

$\text{ b. }1{\mathrm{NH}}_2{\mathrm{O}}_2=5.6\mathrm{V}$

$\therefore 10{\mathrm{NH}}_2{\mathrm{O}}_2=56\mathrm{V}$

$\text{ d. }{\mathrm{N}}_1{\mathrm{V}}_1+{\mathrm{N}}_2{\mathrm{V}}_2={\mathrm{N}}_3{\mathrm{V}}_3\left({\mathrm{V}}_3=10+40=50\mathrm{mL}\right)$

$\begin{array}{l}10\times 10+40\times \frac{5}{8}\times 2={\mathrm{N}}_3\times 50\\ {\mathrm{N}}_3\left({\mathrm{H}}_2{\mathrm{O}}_2\right)=3\mathrm{N}\end{array}$

The volume strength of ${\mathrm{H}}_2{\mathrm{O}}_2=5.6\times 3=16.8\mathrm{V}$