JEE Main 2022 · 29 June · evening · Single correct · +4 / −1
A vessel contains 16g of hydrogen and 128g of oxygen at standard temperature and pressure. The volume of the vessel in cm³ is :
- (a) 72 × 10⁵
- (b) 32 × 10⁵
- (c) 27 × 10⁴
- (d) 54 × 10⁴
Answer and solution
Answer (c) 27 × 10⁴
Solution
Physics sir · 3 steps
Gases in a mixture just add up their moles, and at STP every mole takes the same 22.4 litres.
- Count the moles. Hydrogen and oxygen are diatomic, so use molar masses 2 g mol⁻¹ and 32 g mol⁻¹. That gives 16/2 = 8 mol of H₂ and 128/32 = 4 mol of O₂.
- Add them. For an ideal gas mixture the volume depends only on the total number of moles, not on which gas they belong to. Total n = 8 + 4 = 12 mol.
V = n × 22400 cm³ = 12 × 22400 = 268800 cm³
- So. V = 2.688 × 10⁵ cm³ ≈ 27 × 10⁴ cm³, which is option (c).
The idea to keep. At STP, one mole of any ideal gas occupies 22.4 L = 22400 cm³. For a mixture, find the total moles using the molecular mass of each gas, then multiply.
Answer from the NTA final answer key. Solution by PrepWiser.