The reaction between M\(_2\)CO\(_3\) and HCl is: \[ \text{M}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{MCl} + \text{H}_2\text{O} + \text{CO}_2. \] From the principle of atomic conservation, 1 mole of M\(_2\)CO\(_3\) produces 1 mole of CO\(_2\). Given: \[ \text{Moles of CO}_2 = 0.01 \, \text{mol}. \] \[ \text{Moles of M}_2\text{CO}_3 = 0.01 \, \text{mol}. \] The mass of M\(_2\)CO\(_3\) is 1 g, so: \[ \text{Molar mass of M}_2\text{CO}_3 = \frac{\text{Mass}}{\text{Moles}} = \frac{1}{0.01} = 100 \, \text{g mol}^{-1}. \]
Final Answer: \( \boxed{100} \, \text{g mol}^{-1} \).
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 