Gas Laws & Ideal Gas Equation question from JEE Main 2022 (29 Jun, Shift 2)

Answer: (c)

1. Moles: $n_{H_2} = \frac{16}{2} = 8$, $n_{O_2} = \frac{128}{32} = 4$, total $n = 12$.
2. At STP 1 mol occupies $22.4\text{ L} = 22400\text{ cm}^3$.
3. $V = 12 \times 22400 = 2.69 \times 10^5\text{ cm}^3 \approx 27 \times 10^4\text{ cm}^3$.
4. So the answer is (c) $27 \times 10^4$.