Question

$10$ ml of $NaH{C}_2{O}_4$ solution is neutralised by $10$ ml of $0.1MNaOH$ solution. $10$ ml of same $NaH{C}_2{O}_4$ solution is oxidised by $10$ ml of $KMn{O}_4$ solution in acidic medium. Hence, molarity of $KMn{O}_4$ is :

• A. $0.1M$
• B. $0.2M$
• C. $0.04M$
• D. $0.02M$

Answer 1

Answer: (C)

$0.04M$

The number of moles of $NaH{C}_2{O}_4$ is equal to the number of moles of $NaOH$.
$\therefore 10\times x=10\times 0.1$
$x=0.1M$
$KMn{O}_4$ oxidizes ${C_2{O}_4}^{2-}$
$0.1M{H{C}_2{O}_4}^{-}=0.2M{H{C}_2{O}_4}^{-}$
$10\times 0.2=10\times N^{\prime }KMn{O}_4$
Hence, $N^{\prime }=0.2N$
The oxidation number of manganese changes by $5$.
Hence, $Molarity=\frac{normality}{5}=\frac{0.2}{5}=0.04M$