Question

How much volume of 0.1 M $Zn(Mn{O}_4{)}_2$ is required to react with 50 ml of 0.2 M ferrous oxalate? Given $Mn{O}_4^{-}$ reduces into $M{n}^{+2}$ in acidic medium.

• A. 30 ml
• B. 60 ml
• C. 40 ml
• D. None of these

Answer 1

The correct option is C 40 ml

$\therefore$ $n_1{M}_1{V}_1=n_2{M}_2{V}_2$

$5\times 0.1\times V_1=2\times 0.2\times 50$

$V_1=\frac{2\times 0.2\times 50}{5\times 0.1}$

$V_1=40$ ml