Active mass of 0.64g `SO_(2)` in 10 lit vessel is

Options

A. `10^(-2)` M
B. `10^(-3)`M
C. `10^(-1)` M
D. 0.64g

Answer

To find the active mass of 0.64 g of sulfur dioxide (SO₂) in a 10-liter vessel, we will follow these steps:

### Step 1: Determine the molecular mass of sulfur dioxide (SO₂)
The molecular mass of SO₂ can be calculated as follows:
- Sulfur (S) has an atomic mass of approximately 32 g/mol.
- Oxygen (O) has an atomic mass of approximately 16 g/mol.
- Therefore, the molecular mass of SO₂ = 32 + (2 × 16) = 32 + 32 = 64 g/mol.

### Step 2: Calculate the number of moles of SO₂
To find the number of moles of SO₂, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molecular mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{0.64 \text{ g}}{64 \text{ g/mol}} = 0.01 \text{ moles} \]

### Step 3: Calculate the concentration (active mass) of SO₂
The concentration (active mass) in moles per liter (M) can be calculated using the formula:
\[ \text{Concentration (M)} = \frac{\text{Number of moles}}{\text{Volume (L)}} \]
Substituting the values:
\[ \text{Concentration (M)} = \frac{0.01 \text{ moles}}{10 \text{ L}} = 0.001 \text{ M} = 10^{-3} \text{ M} \]

### Final Answer
The active mass of 0.64 g of SO₂ in a 10-liter vessel is \(10^{-3} \text{ M}\).

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