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A 0.518 g s...

Question

A $0.518 g$ sample of limestone is dissolved in $H C l$ and then the calcium is precipitated as $C a C_{2} O_{4}$ . After filtering and washing the precipitate, it requires $40 m L$ of $0.25 N K M n O_{4}$ solution acidified with $H_{2} S O_{4}$ to titrated it as,
${M n O_{4}}^{-} + H^{+} + {C_{2} O_{4}}^{2 -} \to C O_{2} + M n^{2 +} + 2 H_{2} O$
The percentage of $C a O$ in the sample is:

A

54

%

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B

27.1

%

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C

42

%

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D

84

%

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Solution

The correct option is A $54$ %
Number of milliequivalents of $C a C_{2} O, K M n O_{4}$ and $C a O$ will be same.
$40 \times 0.25 = \frac{W}{56 / 2} \times 1000$
$W = 0.28 g$ (Mass of $C a O$ )
% $C a O = \frac{0.28}{0.5218} \times 100 = 54$ %.

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