The volume of  0.1M permanganate solution required to oxidise 300 ml of 0.1 M thiosulphate solution in faintly alkaline medium to form sulphate ion is 

Mole ratio = 8:3;

$8\stackrel{+7}{\mathrm{Mn}}{\mathrm{O}}_4^{-}+3{\stackrel{+2}{\mathrm{S}}}_2{\mathrm{O}}_3^{-2}+{\mathrm{H}}_2\mathrm{O}\to 8\stackrel{+4}{\mathrm{Mn}}{\mathrm{O}}_2+6\stackrel{+6}{\mathrm{S}}{\mathrm{O}}_4^{-2}+2{\mathrm{OH}}^{-};$

$$\begin{gathered} \frac{{\mathrm{M}}_1{\mathrm{V}}_1}{{\mathrm{n}}_1}=\frac{{\mathrm{M}}_2{\mathrm{V}}_2}{{\mathrm{n}}_2} \\ \frac{0.1\times {\mathrm{V}}_1}{8}=\frac{300\times 0.1}{3} \\ =800 \mathrm{ml}; \\ \mathrm{Alternate} \mathrm{method}; \\ \mathrm{Decrease} \mathrm{in} \mathrm{ox}.\mathrm{no}/\mathrm{mole} \mathrm{of} {\mathrm{KMnO}}_4=3; \\ \mathrm{Increase} \mathrm{in} \mathrm{ox}.\mathrm{no}/\mathrm{mole} \mathrm{of} {\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3=8; \\ {\mathrm{V}}_1{\mathrm{N}}_1={\mathrm{V}}_2{\mathrm{N}}_2; \\ {\mathrm{V}}_1\left(0.1\mathrm{M}\right)3=300\left(0.1\right)8; \\ {\mathrm{V}}_1=800 \mathrm{ml}; \end{gathered}$$