Question

$3$ moles of a mixture of $FeS{O}_4$ and $F{e}_2(S{O}_4{)}_3$ required $100$ mL of $2$M $KMn{O}_4$ solution in acidic medium. Hence, mole fraction of $FeS{O}_4$ in the mixture is:

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{2}{5}$
• D. $\frac{3}{5}$

Answer 1

The correct option is A $\frac{1}{3}$

${Mn{O}_4}^{-}$ oxidises only $FeS{O}_4$.
${Mn{O}_4}^{-}+5F{e}^{2+}\to 5F{e}^{3+}+M{n}^{2+}$
1 mole of $Mn{O_4}^{-}$ oxidises 5 moles of $FeS{O}_4$

So, $\frac{100\times 2}{1000}=0.2$ mol ${Mn{O}_4}^{-}$ oxidises $1$ mol $FeS{O}_4$ in $3$ mol mixture.
Mole fraction of $FeS{O}_4=\frac{1}{3}$.