Question

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: -

• A. 1186
• B. 84.3
• C. 118.6
• D. 11.86

Answer 1

The correct option is B 84.3

Given chemical $e{q}^n$
$M_{2}C{O}_3+2HCl\to 2MCl+H_{2}O+C{O}_{2}1gm$
0.01186mol
$\Rightarrow$ from the balanced chemical $e{q}^n.$
$\frac{1}{M}=0.01186$
$\Rightarrow M=84.3gm/mol$