Question

$3$ mol of a mixture of $FeS{O}_4$ and $F{e}_2(S{O}_4{)}_3$ required $100mL$ of $2M$ $KMn{O}_4$ 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 $\frac{1}{3}$

Here, FeSO4 has Fe in +2 oxidation state and Fe2(SO4)3 has Fe in +3 state.

So, $KMn{O}_4$ will oxidise only Fe${}^{2+}$ to Fe${}^{3+}$

Therefore, the reaction occurring is

$2KMn{O}_4+8H_{2}S{O}_4+10FeS{O}_4\to K_{2}S{O}_4+2MnS{O}_4+5F{e}_2(S{O}_4{)}_3+8H_{2}O$

Equivalents of $FeS{O}_4=$ Equivalents of $KMn{O}_4$

or, Equivalents of $FeS{O}_4=0.1L\times (2\times 5)N$   (Since Normality=Molarity×n−factor and for KMnO4 in acidic medium, n=5)

Therefore, Equivalents of FeSO4 = 1

or, Moles of FeSO4 $=\frac{\text{Equivalents}}{\text{n−factor}}=\frac{1}{1}$ (Since for FeSO4 n=1)

Therefore, Mole fraction of FeSO4 in the mixture$=\frac{{\text{Moles of FeSO}}_4}{\text{Moles of mixture}}$

$=\frac{1}{3}$

Hence, the correct option is $A$