3.12 g of oxygen is adsorbed on 1.2 g of platinum metal. The volume of oxygen adsorbed per gram of the adsorbent at 1 atm and 300 K is L is _____________.
[R = 0.0821 L atm `K^(-1) "mol"^(-1)`]

Answer

To solve the problem, we need to find the volume of oxygen adsorbed per gram of platinum at 1 atm and 300 K. We will use the ideal gas law and the given data to calculate this.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of oxygen (O₂) = 3.12 g
- Mass of platinum (Pt) = 1.2 g
- Temperature (T) = 300 K
- Pressure (P) = 1 atm
- Ideal gas constant (R) = 0.0821 L atm K⁻¹ mol⁻¹

2. **Calculate the Number of Moles of Oxygen:**
- The molar mass of oxygen (O₂) = 32 g/mol.
- Number of moles of oxygen (n) can be calculated using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{3.12 \, \text{g}}{32 \, \text{g/mol}} = 0.0975 \, \text{mol} \]

3. **Use the Ideal Gas Law to Calculate the Volume of Oxygen:**
- The ideal gas law is given by:
\[ PV = nRT \]
- Rearranging for volume (V):
\[ V = \frac{nRT}{P} \]
- Substituting the values:
\[ V = \frac{(0.0975 \, \text{mol}) \times (0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}) \times (300 \, \text{K})}{1 \, \text{atm}} \]
- Calculating the volume:
\[ V = \frac{(0.0975) \times (0.0821) \times (300)}{1} \approx 2.40 \, \text{L} \]

4. **Calculate the Volume of Oxygen Adsorbed per Gram of Platinum:**
- To find the volume of oxygen adsorbed per gram of platinum:
\[ \text{Volume per gram of platinum} = \frac{V}{\text{mass of platinum}} = \frac{2.40 \, \text{L}}{1.2 \, \text{g}} = 2.00 \, \text{L/g} \]

### Final Answer:
The volume of oxygen adsorbed per gram of platinum at 1 atm and 300 K is **2.00 L/g**.