| ⇦ |
| ⇨ |
8.2 L of an ideal gas weight 9.0 gm at 300 K and 1 atm pressure. The molecular mass of the gas is
Options
(a) 9
(b) 18
(c) 27
(d) 36
Correct Answer:
27
Explanation:
PV = nRT = (m / M) RT ⇒ M = mRT / PV = (9.0 × 0.0821 × 300) / (1 × 8.2) = 27
Related Questions: - If the rate of diffusion of CH₄ is twice of that of a gas X, then what is the molecular
- Argon gas was discovered by
- The hydrogen ion concentration in mol/dm, in a 0.2 M solution of a weak acid
- For the equilibrium, H₂O(l) ⇌ H₂O(g) at 1 atm and 298 K,
- At 27⁰C average kinetic energy of one mole of helium gas can be given
Topics: States of Matter Gases and Liquids
(80)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If the rate of diffusion of CH₄ is twice of that of a gas X, then what is the molecular
- Argon gas was discovered by
- The hydrogen ion concentration in mol/dm, in a 0.2 M solution of a weak acid
- For the equilibrium, H₂O(l) ⇌ H₂O(g) at 1 atm and 298 K,
- At 27⁰C average kinetic energy of one mole of helium gas can be given
Topics: States of Matter Gases and Liquids (80)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Leave a Reply