When 10   mL of an aqueous solution of KMnO 4 was titrated in acidic medium, equal volume of 0 . 1 M of…

$\underset{\mathrm{vf}=5}{{\mathrm{MnO}}_4^{-}}+\underset{\mathrm{vf}=1}{{\mathrm{Fe}}^{2+}⟶}{\mathrm{Mn}}^{2+}+{\mathrm{Fe}}^{3+}$

Mili equ. of ${\mathrm{MnO}}_4{}^{-}=$ mili equ. of ${\mathrm{Fe}}^{2+}$

$5[\mathrm{M}\times 10]=1[0.1\times 10]$

$\mathrm{M}=\left[\frac{0.1}{5}\right]=0.02\frac{\text{ mole }}{\text{ Lit }}$

Strength of ${\mathrm{KMnO}}_4=0.02\times 158\frac{\mathrm{gram}}{\mathrm{Lit}}=3.16 \mathrm{g}/\mathrm{L}=316\times {10}^{-2} \mathrm{g}/\mathrm{L}$