Question

$KMn{O}_4$ reacts with ferrous ammonium sulphate according to the equation $Mn{O}_4^{-}+5F{e}^{2+}+8H^{+}\to M{n}^{2+}+5F{e}^{3+}+4H_{2}O$, here 10 ml of 0.1 M $KMn{O}_4$ is equivalent to

• A. 20 ml of 0.1 M $FeS{O}_4$
• B. 30 ml of 0.1 M $FeS{O}_4$
• C. 40 ml of 0.1 M $FeS{O}_4$
• D. 50 ml of 0.1 M $FeS{O}_4$

Answer 1

The correct option is D 50 ml of 0.1 M $FeS{O}_4$

$KMn{O}_4=Mohrsalt$
$\frac{M_1{V}_1}{n_1}=\frac{M_2{V}_2}{n_2};\frac{0.1\times 10}{1}=\frac{M_2{V}_2}{5};M_2{V}_2=5$

The option that has the same number of moles will be the answer.