`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

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