At equilibrium, the amount of HI in a 3 litre vessel was 12.8 g. Its equilibrium concentration is

Options

A. 4.267 M
B. 0.033 M
C. 0.1 M
D. 0.2 M

Explanation

To find the equilibrium concentration of HI in a 3-liter vessel, we will follow these steps:

### Step 1: Calculate the number of moles of HI
To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of HI (Hydrogen Iodide) can be calculated as follows:
- Molar mass of H = 1 g/mol
- Molar mass of I = 127 g/mol
- Therefore, molar mass of HI = 1 + 127 = 128 g/mol

Now, we can calculate the number of moles of HI:
\[ \text{Number of moles of HI} = \frac{12.8 \text{ g}}{128 \text{ g/mol}} = 0.1 \text{ mol} \]

### Step 2: Calculate the concentration (molarity) of HI
Molarity (M) is defined as the number of moles of solute per liter of solution:
\[ \text{Molarity (M)} = \frac{\text{Number of moles}}{\text{Volume of solution in liters}} \]
In this case, the volume of the vessel is 3 liters:
\[ \text{Molarity of HI} = \frac{0.1 \text{ mol}}{3 \text{ L}} = 0.0333 \text{ M} \]

### Final Answer
The equilibrium concentration of HI in the 3-liter vessel is **0.0333 M**.

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