Question

$1\mathrm{gram}$ of a carbonate $\left({\mathrm{M}}_2{\mathrm{CO}}_3\right)$ on treatment with $\mathrm{excess}\mathrm{HCl}$ produces $0.01186\mathrm{moles}$ of ${\mathrm{CO}}_2$. The molar mass of ${\mathrm{M}}_2{\mathrm{CO}}_3$ in $\mathrm{g}{\mathrm{mol}}^{-1}$ is :

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

Answer 1

The correct option is D

$84.3$

Explanation for the correct option:

Step 1: Given data:

Weight of ${\mathrm{M}}_2{\mathrm{CO}}_3$ is ${\mathrm{w}}_{{\mathrm{M}}_2{\mathrm{CO}}_3}=1\mathrm{gram}$

The number of moles of ${\mathrm{CO}}_2$ produced is ${\mathrm{n}}_{{\mathrm{CO}}_2}=0.01186\mathrm{moles}$.

Step 2: Find the number of moles of the unknown compound ${\mathbf{M}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$:

${\mathrm{M}}_2{\mathrm{CO}}_3$ on reaction with $\mathrm{excess}\mathrm{HCl}$ gives ${\mathrm{CO}}_2$ gas

The balanced equation can be written as:

${\mathrm{M}}_2{\mathrm{CO}}_3+2\mathrm{HCl}\to 2\mathrm{MCl}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$

From the above balanced equation,

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{M}}_2{\mathrm{CO}}_3=\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{CO}}_2 \\ \Rightarrow \mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{M}}_2{\mathrm{CO}}_3=0.01186\mathrm{moles} \end{gathered}$$

Step 3: Find the molar mass of an unknown compound ${\mathbf{M}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$:

$$\begin{gathered} \mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{M}}_2{\mathrm{CO}}_3=\frac{\mathrm{Weight}\mathrm{of}{\mathrm{M}}_2{\mathrm{CO}}_3}{\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{M}}_2{\mathrm{CO}}_3} \\ =\frac{1\mathrm{g}}{0.01186\mathrm{mol}} \\ =84.3\mathrm{g}{\mathrm{mol}}^{-1} \end{gathered}$$

Hence, option D is correct. The Molar mass of the unknown compound ${\mathbf{M}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$ is $\mathbf{84}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{g}\mathbf{}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}$.