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Mass of ${KHC}_{2} O_{4}$ (potassium oxalate) required to reduce 100 mL of 0.02 M ${KMnO}_{4}$ in acidic
medium $({to Mn}^{2+})$ is x g and to neutralise 100 mL of 0.05 M $Ca {(OH)}_{2}$ is y g, then

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x = 2y

none is correct

x = y

2x = y

answer is B.

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Detailed Solution

${KHC}_{2} O_{4} {+ KMnO}_{4}$

$n_{eq} {KHC}_{2} O_{4} {= n}_{eq} {KMnO}_{4}$

$\frac{w}{128} × 2 = 100 × 0.02 × 5$

${w = 100 × 0.02 × 5 × 64; w}_{x} {= 640 mg ; n}_{eq} {KHC}_{2} O_{4} {= n}_{eq} Ca {(OH)}_{2}$

$\frac{w}{128} × 1 = 100 × 0.05 × 2$

$w_{y} = 1280 mg$

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