${KMnO}_{4}$ reacts with oxalic acid according to the equation, $2 {MnO}_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 {Mn}^{2 +} + 10 {CO}_{2} + 8 H_{2} O$
here $20 ml$ of $0.1 M {KMnO}_{4}$ is equivalent to

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$20 ml$ of $0.5 {MH}_{2} C_{2} O_{4}$

$50 ml$ of $0.5 {MH}_{2} C_{2} O_{4}$

$50 ml$ of $0.1 {MH}_{2} C_{2} O_{4}$

$20 ml$ of $0.1 {MH}_{2} C_{2} O_{4}$

answer is B.

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Detailed Solution

${KMnO}_{4}$ Oxalic acid

$\frac{M_{1} V_{1}}{n_{1}} = \frac{M_{2} V_{2}}{n_{2}} : \frac{20 \times 0 .1}{2} = \frac{M_{2} V_{2}}{5}; M_{2} V_{2} = 5$

for (b) $M_{1} V_{1} = 50 \times 0 .1 = 5$

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