`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. 11.86
B. 1186
C. 84.3
D. 118.6

Explanation

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

### Step 1: Understand the reaction
The reaction of the carbonate \( M_2CO_3 \) with hydrochloric acid (HCl) produces carbon dioxide (CO₂), metal chloride (MCl), and water (H₂O). The balanced reaction can be written as:
\[ M_2CO_3 + 6HCl \rightarrow 2MCl + H_2O + CO_2 \]
From this reaction, we can see that 1 mole of \( M_2CO_3 \) produces 1 mole of CO₂.

### Step 2: Relate moles of CO₂ to moles of \( M_2CO_3 \)
Given that 0.01186 moles of CO₂ are produced, it follows that:
\[ \text{Moles of } M_2CO_3 = \text{Moles of CO₂} = 0.01186 \text{ moles} \]

### Step 3: Use the formula for moles
The number of moles can be calculated using the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
Here, the given mass of \( M_2CO_3 \) is 1 gram.

### Step 4: Set up the equation
Substituting the known values into the formula gives:
\[ 0.01186 = \frac{1 \text{ g}}{M} \]
Where \( M \) is the molar mass of \( M_2CO_3 \).

### Step 5: Rearranging the equation to solve for \( M \)
Rearranging the equation to find \( M \):
\[ M = \frac{1 \text{ g}}{0.01186} \]

### Step 6: Calculate the molar mass
Now, performing the calculation:
\[ M = \frac{1}{0.01186} \approx 84.3 \text{ g/mol} \]

### Conclusion
Thus, the molar mass of \( M_2CO_3 \) is approximately \( 84.3 \text{ g/mol} \).