CATES OF MATTER 1. Hydrogen, \( \mathrm{H}_{2} \) and oxygen. \( \mathrm{O}_{2} \) gases are placed in a container at the same temperature. Using Graham's law of diffusion, calculate the relative rate of difrusion of hydrogen compared to oxygen.

To calculate the relative rate of diffusion of hydrogen (\(\mathrm{H}_2\)) compared to oxygen (\(\mathrm{O}_2\)) using **Graham's Law of Diffusion**, follow these steps: ### **Graham's Law of Diffusion** Graham's Law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it is expressed as: \[ \frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} \] where: - \(\text{Rate}_1\) and \(\text{Rate}_2\) are the rates of diffusion of gas 1 and gas 2, respectively. - \(M_1\) and \(M_2\) are the molar masses of gas 1 and gas 2, respectively. ### **Given Data** - **Hydrogen (\(\mathrm{H}_2\)):** - Molar mass (\(M_1\)) = 2 g/mol - **Oxygen (\(\mathrm{O}_2\)):** - Molar mass (\(M_2\)) = 32 g/mol ### **Calculating the Relative Rate** \[ \frac{\text{Rate}_{\mathrm{H}_2}}{\text{Rate}_{\mathrm{O}_2}} = \sqrt{\frac{M_{\mathrm{O}_2}}{M_{\mathrm{H}_2}}} = \sqrt{\frac{32\ \text{g/mol}}{2\ \text{g/mol}}} = \sqrt{16} = 4 \] ### **Conclusion** Hydrogen gas (\(\mathrm{H}_2\)) diffuses **4 times faster** than oxygen gas (\(\mathrm{O}_2\)) under the same conditions of temperature and pressure. **Answer:** Hydrogen gas diffuses four times as rapidly as oxygen under the same conditions.