Question

What volume of $0.1M$ $KMn{O}_4$ is needed to oxidise $5$ mg of ferrous oxalate in acidic medium? (Molecular weight of ferrous oxalate is $144$ g/mol)

• A. $0.20$ mL
• B. $0.1$ mL
• C. $0.4$ mL
• D. $2.08$ mL

Answer 1

The correct option is A $0.20$ mL

Milliequivalents of $KMn{O}_4=V\times 0.1\times 5$

Milliequivalents of $Fe(C_2{O}_4{)}_2=\frac{\text{Weight in g}}{\text{Ew}}\times 1000$

$\left[\begin{array}{c}F{e}^{+2}\to F{e}^{+3}+e^{-}\\ C_2{O}_4^{2-}\to 2C{O}_2+2e^{-}\end{array}\right](n=3)$

Equivalent weight of $Fe\left(C_2{O}_4\right)=\frac{Mw}{3}=\frac{144}{3}$

$\therefore$ Milliequivalents of $Fe\left(C_2{O}_4\right)=\frac{5\times {10}^{-3}}{\frac{144}{3}}\times {10}^3=\frac{5\times 3}{144}$

$\therefore$ Milliequivalents of $KMn{O}_4=$ Milliequivalents of $Fe\left(C_2{O}_4\right)$

$\therefore V\times 0.1\times 5=\frac{5\times 3}{144}$

$\therefore V=0.20$ mL