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 in L is ______
[R = 0.0821 L atm K$$^{-1}$$ mol$$^{-1}$$]

We are given that 3.12 g of oxygen is adsorbed on 1.2 g of platinum metal. We need to find the volume of oxygen adsorbed per gram of the adsorbent at 1 atm and 300 K.

First, find the moles of $$O_2$$ adsorbed per gram of Pt:

$$\text{Moles of } O_2 = \frac{3.12}{32} = 0.0975 \text{ mol}$$

$$\text{Moles of } O_2 \text{ per gram of Pt} = \frac{0.0975}{1.2} = 0.08125 \text{ mol/g}$$

Using the ideal gas law to find the volume at 1 atm and 300 K:

$$V = \frac{nRT}{P} = \frac{0.08125 \times 0.0821 \times 300}{1}$$

$$V = 0.08125 \times 24.63 = 2.0012 \text{ L}$$

$$V \approx 2 \text{ L}$$

The answer is $$\boxed{2}$$.