$1 \mathrm{~g}$ of a carbonate $\left(\mathrm{M}_{2} \mathrm{CO}_{3}\right)$ on treatment with excess $\mathrm{HCl}$ produces $0.01 \mathrm{~mol}$ of $\mathrm{CO}_{2}$. The molar mass of $\mathrm{M}_{2} \mathrm{CO}_{3}$ is __________ $\mathrm{g} ~\mathrm{mol}^{-1}$. (Nearest integer)
Answer (integer)
100
Solution
When the carbonate reacts with HCl, it produces CO<sub>2</sub>, as shown in the following balanced equation:
<br/><br/>
$$\mathrm{M}_2\mathrm{CO}_3 + 2\mathrm{HCl} \rightarrow 2\mathrm{MCl} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2$$
<br/><br/>
From the problem, we know that 1 g of the carbonate produces 0.01 mol of CO<sub>2</sub>. <br/><br/>To determine the molar mass of the carbonate, we can use the stoichiometry of the reaction:
<br/><br/>
1 mol of $\mathrm{M}_2\mathrm{CO}_3$ produces 1 mol of CO2.
<br/><br/>
So, 0.01 mol of CO<sub>2</sub> corresponds to 0.01 mol of $\mathrm{M}_2\mathrm{CO}_3$.
<br/><br/>
Now, we can find the molar mass of $\mathrm{M}_2\mathrm{CO}_3$:
<br/><br/>
$$\frac{\text{mass of }\mathrm{M}_2\mathrm{CO}_3}{\text{moles of }\mathrm{M}_2\mathrm{CO}_3} = \text{molar mass of }\mathrm{M}_2\mathrm{CO}_3$$
<br/><br/>
$\frac{1 \text{ g}}{0.01 \text{ mol}} = 100 \text{ g mol}^{-1}$
<br/><br/>
So, the molar mass of $\mathrm{M}_2\mathrm{CO}_3$ is approximately 100 g/mol.
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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