Question

1 gram of a carbonate $\left(M_{2}C{O}_3\right)$ on treatment with excess $HCl$ produces 0.01186 moles of $C{O}_2$. The molar mass of $\left(M_{2}C{O}_3\right)$ in $gmo{l}^{-1}$ is:

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

Answer 1

The correct option is A 84.3

$M_{2}C{O}_3+2HCl\to C{O}_2+2MCl+H_{2}O$

Moles of $M_{2}C{O}_3$ = Moles of $C{O}_2$ produced.

Moles of $M_{2}C{O}_3=\frac{w}{molarmass}=0.01186$

Weight of the carbonate $\left(w\right)$ = 1 $g$

Moles of $M_{2}C{O}_3=\frac{1}{molarmass}=0.01186$

$\therefore$ Molar mass $=84.3gmo{l}^{-1}$