Correct Answer - D
Sample of limestone is only 90 % pure.
Therefore, 100 g limestone sample contains 90 g pure
`:.` 10 g limestone sample will contain `(10xx90)/(100)`
`9.00 g` pure
As limestone is nothing but `CaCO_(3)`
`:.` 10 g limestone sample will contain `(10xx90)/(100) = 9.00` g pure
As limestone is nothing but `CaCO_(3)`
`:.` The molecular mass of `CaCO_(3) = 100`
100 g of limestone gives 22.4 litre of `CO_(2)` at S.T.P.
`:.` 9 g of limestone will give
`(22.4xx9)/(100) = 2.016` litre at S.T.P.
Hence, 10g impure sample will give 2.016 litre of `CO_(2)` at STP