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

Options

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

Answer

B

Explanation

Given chemical `"eq"^(n)`
`underset(1 gm)(M_(2)CO_(3))+2HCl to 2MCl +H_(2)O+underset(0.01186)(CO_(2))`
From the above chemical `"eq"^(n)`
`nM_(2)CO_(3)=nCO_(2)`
`(1)/("Molar mass of" M_(2)CO_(3))=0.01186`
`:. "Molar mass of" M_(2)CO_(3)=(1)/(0.01186)`
`M=84.3 g // mol`