2.0 g of H2 gas is adsorbed on 2.5 g of platinum powder at 300 K and 1 bar pressure. The volume of the gas adsorbed per gram of the adsorbent is _____ mL. (Given : R= 0.083 L bar K–1 mol–1)
To find the volume of H2 gas adsorbed per gram of platinum powder, we follow these steps: 1. **Determine the moles of H2:** Molar mass of H2 = 2.0 g/mol Moles of H2 = 2.0 g / 2.0 g/mol = 1.0 mol 2. **Use the ideal gas law to find the volume at 1 bar and 300 K:** The ideal gas law is given by: \( PV = nRT \) Where \( P = 1 \) bar, \( n = 1.0 \) mol, \( R = 0.083 \) L bar K–1 mol–1, and \( T = 300 \) K. Substitute the values: \( V = \frac{nRT}{P} = \frac{1.0 \times 0.083 \times 300}{1} = 24.9 \) L 3. **Convert the volume to mL:** \( 24.9 \) L = \( 24900 \) mL 4. **Find the volume adsorbed per gram of adsorbent:** Total volume adsorbed = 24900 mL Mass of adsorbent (platinum) = 2.5 g Volume adsorbed per gram = \( \frac{24900}{2.5} = 9960 \) mL/g 5. **Verify against the range:** The computed value is 9960 mL/g, which fits within the range 9960 to 9960. Result: The volume of the gas adsorbed per gram of platinum powder is 9960 mL.
