`1g` of a carbonate `(M_(2)CO_(3))` on treatment with excess `HCl` produces `0.01186` mole of `CO_(2)`. The molar mass of `M_(2)CO_(3)` in `g mol^(-1)` is

Options

A. (a) `118.6`
B. (b) `11.86`
C. (c) `84.3`
D. (d) `72.4`

Answer

To find the molar mass of the carbonate \( M_2CO_3 \), we can follow these steps:

### Step 1: Understand the Reaction
When \( M_2CO_3 \) reacts with excess hydrochloric acid (HCl), it produces carbon dioxide (CO2), water (H2O), and metal chloride (MCl). The balanced reaction can be represented as:
\[ M_2CO_3 + 2HCl \rightarrow 2MCl + H_2O + CO_2 \]
From this reaction, we can see that 1 mole of \( M_2CO_3 \) produces 1 mole of \( CO_2 \).

### Step 2: Relate Moles of CO2 to Moles of M2CO3
The problem states that 1 gram of \( M_2CO_3 \) produces 0.01186 moles of \( CO_2 \). Since the stoichiometry of the reaction shows that 1 mole of \( M_2CO_3 \) produces 1 mole of \( CO_2 \), we can conclude that:
\[ \text{Moles of } M_2CO_3 = 0.01186 \text{ moles} \]

### Step 3: Use the Molar Mass Formula
The molar mass (M) can be calculated using the formula:
\[ \text{Moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
Rearranging this gives:
\[ \text{Molar mass} = \frac{\text{Given mass}}{\text{Moles}} \]

### Step 4: Substitute the Values
Here, the given mass of \( M_2CO_3 \) is 1 gram and the moles calculated is 0.01186 moles. Therefore:
\[ \text{Molar mass} = \frac{1 \text{ g}}{0.01186 \text{ moles}} \approx 84.3 \text{ g/mol} \]

### Step 5: Conclusion
The molar mass of \( M_2CO_3 \) is approximately \( 84.3 \text{ g/mol} \).

### Final Answer
The molar mass of \( M_2CO_3 \) is **84.3 g/mol**.

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