The active mass of `64 g` of `HI` in a `2-L` flask would be

Options

A. `2`
B. `1`
C. `5`
D. `0.25`

Explanation

To find the active mass of `64 g` of `HI` in a `2-L` flask, we can follow these steps:

### Step 1: Determine the molecular weight of HI
- The molecular weight of HI can be calculated by adding the atomic weights of hydrogen (H) and iodine (I).
- Hydrogen (H) has an atomic weight of approximately `1 g/mol`.
- Iodine (I) has an atomic weight of approximately `127 g/mol`.
- Therefore, the molecular weight of HI = 1 + 127 = `128 g/mol`.

### Step 2: Calculate the number of moles of HI
- To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molecular weight (g/mol)}} \]
- Substituting the values we have:
\[ \text{Number of moles} = \frac{64 \text{ g}}{128 \text{ g/mol}} = 0.5 \text{ moles} \]

### Step 3: Calculate the concentration of HI
- Concentration (C) is defined as the number of moles of solute per liter of solution. The formula for concentration is:
\[ C = \frac{\text{Number of moles}}{\text{Volume (L)}} \]
- We have `0.5 moles` of HI and the volume of the flask is `2 L`. Therefore:
\[ C = \frac{0.5 \text{ moles}}{2 \text{ L}} = 0.25 \text{ moles/L} \]

### Conclusion
- The active mass (or concentration) of `64 g` of `HI` in a `2-L` flask is `0.25 moles/L`.

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