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. 1186
B. 84.3
C. 118.6
D. 11.86

Answer

B

Explanation

The reaction is :
`M_(2)CO_(3)+ 2HCl to 2Mcl + H_(2)O + CO_(2)`
Number of moles of `CO_(2)` = Number of moles of `M_(2)CO_(3)`
`0.01186 = ("Mass")/("Molar mass of" M_(2)CO_(3))= (1)/("Molar mass of"M_(2)CO_(3))`
`therefore` Molar mass of `M_(2)CO_(3) = (1)/(0.01186) = 84.3 g mol^(-1)`