3.2 g of oxygen (At. wt. = 16) and 0.2 g of hydrogen (At. wt. = 1) are placed in a 1.12 litre flask — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

I need to find the total pressure of a gas mixture of oxygen and hydrogen at 0°C in a 1.12 L flask. **Step 1: Calculate moles of each gas** For oxygen: $$n_{O_2} = \frac{3.2 \text{ g}}{32 \text{ g/mol}} = 0.1 \text{ mol}$$ For hydrogen: $$n_{H_2} = \frac{0.2 \text{ g}}{2 \text{ g/mol}} = 0.1 \text{ mol}$$ **Step 2: Calculate total moles** $$n_{total} = 0.1 + 0.1 = 0.2 \text{ mol}$$ **Step 3: Apply ideal gas law** Using $PV = nRT$ where: - $n = 0.2$ mol - $V = 1.12$ L - $T = 0°C = 273$ K - $R = 0.0821$ L·atm/(mol·K) $$P = \frac{nRT}{V} = \frac{0.2 \times 0.0821 \times 273}{1.12}$$ $$P = \frac{4.48}{1.12} = 4 \text{ atm}$$ Alternatively, using $R = 8.314$ J/(mol·K) gives pressure in Pa, which converts to approximately **4 atm** or **304 kPa** or **405 kPa** depending on the answer choices provided. **Answer: The total pressure is 4 atm** (or equivalent in other units like 405 kPa if using standard conditions where 1 atm ≈ 101.325 kPa). The key principle is that partial pressures are independent (Dalton's Law), so we can simply add total moles and use the ideal gas equation once.