Question

A solution contains $2.52$ g/L of a reductant. $25$ mL of this solution required $20$ mL of $0.01$ M $KMn{O}_4$ in acid medium for oxidation. Find the molar mass of reductant. Given that each of the two atoms, which undergo oxidation per molecule of reductant suffer an increase in oxidation state by one unit.

• A. $126$
• B. $63$
• C. $72$
• D. $144$

Answer 1

The correct option is A $126$

The redox changes are as follows:
$M{n}^{7+}+5e⟶M{n}^{2+}$
$(A^{a+}{)}_2⟶2A{b}^{+}+2e^{-}$
$\therefore$ Eq. mass of reductant $=\frac{M}{2}$
$\because$ Meq. of reductant used in 25 mL$=$ Meq. of $KMn{O}_4=20\times 0.01\times 5$
$\therefore$ Meq. of reductant in 1 litre$=20\times 0.01\times 5\times \frac{1000}{25}=40$
or, $\left(\frac{w}{\frac{M}{2}}\right)\times 1000=40$
or, $\frac{(2.52\times 2\times 1000)}{M}=40$
$\therefore M=126$
Therefore, the molecular mass of reductant is $126$ g/mol.