Question

1 gram of a carbonate $\left(M_2{CO}_3\right)$ on treatment with excess $HCl$ produces 0.01186 mole of ${CO}_2$ the molar mass of $M_2{CO}_3$ in g mole${}^{-1}$ is:

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

Answer 1

Answer: (B)

84.3

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

Here, Moles of $M_{2}C{O}_3$ =Moles of $C{O}_2$

$\frac{1}{M}=0.01186$

$M=\frac{1}{0.01186}$

$M=84.3gmol$${}^{-1}$

Therefore, Molar mass of $M_{2}C{O}_3$ is $84.3gmole$${}^{-1}$