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A `0.518 g` sample of limestone is dissolved in `HCl` and then the calcium is precipitated as `CaC_(2)O_(4)`. After filtering and washing the precipitate, it requires `40.0` filtering and washing the precipitate, it requires `40.0 mL` of `0.250 N KMnO_(4)`, solution acidified with `H_(2)SO_(4)` to titrate it as. The percentage fo `CaO` in the sample is:
`MnO_(4)^(-)+H^(+)+C_(2)O_(4)^(2-)rarrMn^(2+)+CO_(2)+2H_(2)O`
A. `54.0%`
B. `27.1%`
C. `42%`
D. `84%`

1 Answer

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Correct Answer - A
`"Meq. of limestone" = "Meq. Of" CaC_(2)O_(4)`
`= "Meq. of" KMnO_(4) = "Meq. of CaO"`
`:. 40 xx 0.250 = (w)/(56//2) xx 1000`
`:. w_(CaO) = 0.28`
`:. %` of `CaO = (0.28 xx 100)/(0.518) = 54%`

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