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

Explanation

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 hydrochloric acid (HCl), it produces carbon dioxide (CO2), metal chloride (MCl), and water (H2O). The balanced reaction can be represented as:
\[ M_2CO_3 + 6HCl \rightarrow 2MCl + H_2O + CO_2 \]

### Step 2: Determine moles of CO2 produced
According to the problem, 1 g of \( M_2CO_3 \) produces 0.01186 moles of CO2.

### Step 3: Relate moles of CO2 to moles of \( M_2CO_3 \)
From the balanced equation, we see that 1 mole of \( M_2CO_3 \) produces 1 mole of CO2. Therefore, the moles of \( M_2CO_3 \) that reacted is also 0.01186 moles.

### Step 4: Calculate the molar mass of \( M_2CO_3 \)
The molar mass (M) can be calculated using the formula:
\[ \text{Molar Mass} = \frac{\text{mass of the sample}}{\text{number of moles}} \]
Substituting the values we have:
\[ \text{Molar Mass} = \frac{1 \text{ g}}{0.01186 \text{ moles}} \approx 84.3 \text{ g/mol} \]

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

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