Question

63.5 kg sample of $K_2{C}_2{O}_4.3H_2{C}_2{O}_4.4H_{2}O$ is oxidised by 10 moles of $KMn{O}_4$ in acidic medium. Calculate % purity of $K_2{C}_2{O}_4.3H_2{C}_2{O}_4.4H_{2}O$ in the given sample.

Assume that impurity does not react with $KMn{O}_4.$

[Molar mass of $K_2{C}_2{O}_4.3H_2{C}_2{O}_4.4H_{2}O$ is 508]

• A. 5
• B. 8
• C. 9
• D. 3

Answer 1

Answer: (A)

5

${2Mn{O}_4}^{-}+{5C_2{O}_4}^{2-}+16H^{+}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$.

Thus, 2 moles of $KMn{O}_4$ are required for the oxidation of 5 moles of oxalate ions.

1 mole of $K_2{C}_2{O}_4.3H_2{C}_2{O}_4.4H_{2}O$ contains 4 moles of oxalate ions and will require $\frac{2}{5}\times 4=1.6$ moles of $KMn{O}_4$ .

Hence, 10 moles of $KMn{O}_4$ will oxidize $\frac{10}{1.6}=6.25$ moles of $K_2{C}_2{O}_4.3H_2{C}_2{O}_4.4H_{2}O$.

The number of moles present in 63.5 kg of $K_2{C}_2{O}_4.3H_2{C}_2{O}_4.4H_{2}O$ is $\frac{63.5\times 1000}{508}=125$ moles.

The percentage purity is $\frac{6.25}{125}\times 100=5$%