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A 0.518 g sample of limestone is dissolved in HCI and then the calcium is precipitated as CaC₂O₄. After filtering and washing the precipitate, it requires 40.0 mL of 0.250 N KMnO₄ solution acidified with H₂SO₄ to titrate it as. The percentage of CaO in the sample is:
${MnO}_{4}^{-} + H^{+} + C_{2} O_{4}^{2 -} ⟶ {Mn}^{2 +} + {CO}_{2} + 2 H_{2} O$

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answer is A.

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Detailed Solution

$\begin{array}{l} M eq. of {KMnO}_{4} = M eq. of C_{2} O_{4}^{- 2} = M eq. of {CaCO}_{3} \\ 40 \times .25 = M eq. of CaO \\ \frac{10 \times 10^{- 3}}{2} = Mole of CaO \\ % CaO = \frac{5 \times 10^{- 3} \times 56 \times 100}{.518} = 54 % \end{array}$

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