A 0.518 g sample of lime stone is dissolved in HCl and then the calcium is precipitated as `CaC_2O_4` After filtering and washing the precipitate, it requires 40.0 mL of 0.250 N `KMnO_4` solution acidified with `H_2SO_4` to titrate it as, `MnO_(4)^(-)+H^(+) +C_(2)O_(4)^(2-) rarr Mn^(2+)+CO_(2)+2H_(2)O`
The percentage of CaO in the sample is:

Options

A. ` 54.0% `
B. `27.1% `
C. ` 42%`
D. ` 84% `

Answer

A

Explanation

No of milli equivalents of lime stone
= no of milli equivalents of `CaC_(2)O_4`
= no of milli equivalents of `KMnO_4`
`= 40xx0.25 =10`
No of milli moles of lime = 5
`:.` weight of `CaO = 5 xx 56 xx 10^(-3) = 0.28g`
percentage of `CaO= (0.28)/(0.518)xx100=54.05%`