10 g of `MgCO_(3)` decomposes on heating to 0.1 mole `CO_(2)` and 4g MgO. The percent purity of `MgCO_(3)` is (Given that atomic weights of Mg, C and O are 24,12 and 16 u)

Options

A. `24%`
B. `44%`
C. `54%`
D. `84%`

Answer

D

Explanation

`underset(10 g)(MgCO_(3)) overset("Heat")to underset(4 g)(MgO) + underset(0.1"mol")(CO_(2))`
Molar mass of `MgCO_(3) = 24 + 12 + 3 xx 16`
`=84 g"mol"^(-1)`
Molar mass of MgO =24 + 16 = 40 g `"mol"^(-1)`
40 g of MgO is obtained from 84 g of `MgCO_(3)`
4g of MgO will be obtained from `=84/40 xx 4`
`=8.4` g of `MgCO_(3)`
Percentage purity `=8.4/10 xx 100 = 84%`