A 0.1097 sample of ${\mathrm{As}}_2{\mathrm{O}}_3\mathrm{required}26.10\mathrm{mL}\mathrm{of}{\mathrm{KMnO}}_4$ solution for its titration. The molarity of ${\mathrm{KMnO}}_4$ solution is

$\begin{array}{l}\mathrm{Let}\mathrm{molarity}\mathrm{of}{\mathrm{KMnO}}_4\mathrm{solution}\mathrm{be}\mathrm{M}\\ \therefore \mathrm{Eq}.\mathrm{of}{\mathrm{As}}_2{\mathrm{O}}_3=\mathrm{Eq}.{\mathrm{KMnO}}_4\mathrm{solution}\\ {\left(\frac{\mathrm{W}}{\mathrm{E}}\right)}_{{\mathrm{As}}_2{\mathrm{O}}_3} ={\left( \frac{\mathrm{NV}}{1000}\right)}_{{\mathrm{KMnO}}_4}\\ \frac{0.1097}{198}\times 4=\frac{26.10\times \mathrm{M}\times 5}{1000}\\ \left(\mathrm{Equivalent}\mathrm{weight}{\mathrm{As}}_2{\mathrm{O}}_3=\frac{198}{4}{\text{ ; Normality of KMnO}}_4\text{ =M ×5}\right)\\ \mathrm{molarity} \mathrm{of} {\mathrm{KMnO}}_4=0.017\approx 0.018\end{array}$

Note:  n-factor = change in oxidation state