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

# Solution **Step 1: Calculate moles of each gas** For oxygen: $n_{O_2} = \frac{3.2}{32} = 0.1 \text{ mol}$ For hydrogen: $n_{H_2} = \frac{0.2}{2} = 0.1 \text{ mol}$ Total moles: $n_{total} = 0.1 + 0.1 = 0.2 \text{ mol}$ **Step 2: Apply ideal gas law** $$PV = nRT$$ $$P = \frac{nRT}{V}$$ Where: - $n = 0.2$ mol - $R = 0.0821$ L·atm/(mol·K) or $8.314$ J/(mol·K) - $T = 0°C = 273$ K - $V = 1.12$ L **Step 3: Calculate pressure** $$P = \frac{0.2 \times 0.0821 \times 273}{1.12}$$ $$P = \frac{4.48}{1.12} = 4 \text{ atm}$$ Alternatively, using $R = 8.314$ J/(mol·K): $$P = \frac{0.2 \times 8.314 \times 273}{1.12} = \frac{454}{1.12} ≈ 405 \text{ kPa} = 4 \text{ atm}$$ **Answer: The total pressure is 4 atm** (Option B)