1.

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

Answer»

118.6
11.86
1186
84.3

Solution :`underset("1 mole")(M_(2)CO_(3))+2HClrarr 2MCl+H_(2)O+underset("1 mole")(CO_(2))`
`0.01186` mole of `CO_(2)` is produced from `M_(2)CO_(3)`
`"= 1 g(Given)"`
`THEREFORE"1 mole of "CO_(2)" will be produced form "M_(2)CO_(3)`
`=(1)/(0.01186)=84.3g`
But 1 mole of `CO_(2)` is produced from 1 mole of `M_(2)CO_(3)`
`therefore" Molar mass of "M_(2)CO_(3)=84.3"g mol"^(-1)`


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