Question

Acidified $\text{KMn}{{\text{O}}_{\text{4}}}$ oxidises oxalic acid to $\text{C}{{\text{O}}_{2}}.$ What is the volume (in litres) of ${{10}^{-4}}\text{M}\,\text{KMn}{{\text{O}}_{\text{4}}}$ required to completely oxidise 0.5 L of ${{10}^{-2}}\text{M}$ oxalic acid in acid medium?

• A. 125
• B. 1250
• C. 200
• D. 20

Answer 1

Answer: (D)

20

Answer 2

The correct answer is D:20
$\text{KMn}{{\text{O}}_{\text{4}}}$ reacts with oxalic acid according to the following equation.
$2MnO_{4}^{-}+5{{C}_{2}}O_{4}^{2-}+16{{H}^{+}}\to 2M{{n}^{2+}}$ $+10C{{O}_{2}}+8{{H}_{2}}O$
E mass of $KMn{{O}_{4}}=\frac{\text{mol}\text{.mass}}{7-2}$ ${{N}_{KMn{{O}_{4}}}}=5\times molarity=5\times {{10}^{-4}}$
E mass of ${{C}_{2}}O_{4}^{2-}=\frac{\text{mol}\text{.mass}}{2(4-3)}=\frac{\text{mol}\text{.mass}}{2}$
${N}_{C{_{2}}O_{4}^{2-}}=2\times molarity$ $=2\times {{10}^{-2}}$
${{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}$
$5\times {{10}^{-4}}\times {{V}_{1}}=2\times {{10}^{-2}}\times 0.5$
${{V}_{1}}=\frac{2\times {{10}^{-2}}\times 0.5}{5\times {{10}^{-4}}}=20L$