1g of a carbonate ( M _ 2 CO _ 3 ) on treatment with excess HCl...

From Balanced chemical reaction,

${M}_{2}{CO}_{3} + 2HCl \to 2MCl + {H}_{2}O + {CO}_{2}$

Here from this, we get:

$0 . 01mol{CO}_{2} ≡ 0 . 01mol{M}_{2}{CO}_{3} ≡ 1g{M}_{2}{CO}_{3}$

We know that,

Number of moles = $\dfrac{Givenmass}{Molarmass}$

$⇒$Molar mass of ${M}_{2}{CO}_{3} = \dfrac{Givenmassof{M}_{2}{CO}_{3}}{NumberofMolesof{M}_{2}{CO}_{3}} = \dfrac{1}{0 . 01} = 100g / mol$