1 gram of a carbonate `(M_(2)CO_(3))` on treatment with excess HCl produces 0.01186 mol of `CO_(2)`. The molar mass of `M_(2)CO_(3)` in `g*mol^(-1)` is-

Options

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

Answer

D

Explanation

`underset(1" mol")(M_(2)CO_(3)) +2HCl to 2MCl +H_(2)O+underset("1 mol")(CO_(2))`
No. of moles of `M_(2)CO_(3)=("mass")/("molar mass")`
`=(1.0)/("molar mass of "M_(2)CO_(3))`
=0.01186
or, molar mass of `M_(2)CO_(3)=(1.0)/(0.01186)=84.3g*mol^(-1)`.