To measure the quantity of ${\mathrm{MnCl}}_2$ dissolved in an aqueous solution, it was completely converted to ${\mathrm{KMnO}}_4$ using the reaction,
${\mathrm{MnCl}}_2+{\mathrm{K}}_2{\mathrm{S}}_2{\mathrm{O}}_8+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{KMnO}}_4+{\mathrm{H}}_2{\mathrm{SO}}_4+\mathrm{HCl}$ (equation not balanced). Few drops of concentrated $\mathrm{HCl}$ were added to this solution and gently warmed. Further, oxalic acid $\left(225 \mathrm{mg}\right)$) was added in portions till the colour of the permanganate ion disappeared. What is the quantity of ${\mathrm{MnCl}}_2$ (in mg) present in the initial solution? (Atomic weights in $\mathrm{g}\mathrm{ }{\mathrm{mol}}^{-1}:\mathrm{ }\mathrm{Mn}=55, \mathrm{Cl}=35.5$)

Solution

${\mathrm{C}}_2{\mathrm{O}}_4^{--}+{\mathrm{MNO}}_4^{-}\stackrel{{\mathrm{H}}^{+}}{\to }{\mathrm{CO}}_2$
${\mathrm{m}}_{\mathrm{eq}}\mathrm{ }\mathrm{of}\mathrm{ }{\mathrm{C}}_2{\mathrm{O}}_4^{--}={\mathrm{m}}_{\mathrm{eq}}\mathrm{ }\mathrm{of}\mathrm{ }{\mathrm{MNO}}_4^{-}$
$$\begin{gathered} 2\times \frac{0.225}{90}=\mathrm{a}\times 5 \\ \Rightarrow \mathrm{a} =0.001 \mathrm{mole} =1 \mathrm{milimole} \end{gathered}$$
$\mathrm{Mass} \mathrm{of} {\mathrm{MnCl}}_2=1\times \left[55+71\right]$
$=126\mathrm{ }\mathrm{mg}$

Hence, the correct answer is 126.