1g of a carbonate (M2CO3) on treatment with excess HCl produces 0.01 mol of CO2 .The molar mass of M2CO3 is g mol-1. (Nearest integer)
Correct Answer: 100
Solution and Explanation
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: \( \boxed{100} \, \text{g mol}^{-1} \).
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