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

Answer

A

Explanation

`{:(M_(2)CO_(3) + 2HCl,rarr,MCl_(2) + H_(2)O + CO_(2)),((1)/(M_(0))"Mole",,0.01186 mol):}`
`M_(0)=` Molar mass of `M_(2)CO_(3) (M_(2)CO_(3))`
`(1)/(M_(0)) = 0.01186`
`M_(0) = 84.3 g//mol`