1 gram of a carbonate $$\left( {{M_2}C{O_3}} \right)$$  on treatment with excess $$HCl$$  produces 0.01186 mole of $$C{O_2}.$$ The molar mass of $${{M_2}C{O_3}}$$  in $$g\,mo{l^{ - 1}}$$  is:

Solution :
$$\eqalign{ & {\text{Given chemical}}\,e{q^n} \cr & {M_2}C{O_3} + 2HCl \to 2MCl + {H_2}O + C{O_2} \cr & 1gm\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.01186\,mol \cr & {\text{From the above chemical}}\,e{q^n}. \cr & n{M_2}C{O_3} = nC{O_2} \cr & \frac{1}{{{\text{Molar mass of}}\,\,{M_2}C{O_3}}} = 0.01186 \cr & \therefore \,\,{\text{Molar}}\,{\text{mass}}\,{\text{of}}\,{M_2}C{O_3} = \frac{1}{{0.01186}} \cr & M = 84.3\,gm/mol \cr} $$