1 gram of carbonate `(M_(2)CO_(3))` on treatment with excess HCl produces 0.1186 mole of `CO_(2)`. The molar mass of `M_(2)CO_(3) "in g mol"^(-1)`

Options

A. 118.6
B. 11.86
C. 1186
D. 84.3

Answer

D

Explanation

`underset("1 mole")(M_(2)CO_(3))+2HClrarr 2MCl+H_(2)O+underset("1 mole")(CO_(2))`
`0.01186` mole of `CO_(2)` is produced from `M_(2)CO_(3)`
`" = 1 g (Given)"`
`therefore" 1 mole of "CO_(2)" will be produced form "M_(2)CO_(3)`
`=(1)/(0.01186)=84.3g`
But 1 mole of `CO_(2)` is produced from 1 mole of `M_(2)CO_(3)`
`therefore" Molar mass of "M_(2)CO_(3)=84.3"g mol"^(-1)`