Q53.for a real gas at 25 °C temperature and high pressure ( 99 bar) the value of compressibility factor is 2 , so the value of Van 1der Waal's constant ' b' should be_____ ×10−2 L mol−1 (Given R = 0. 083 L bar K−1 mol−1 )
What This Question Tests
This question requires applying VSEPR theory to determine the geometry of multiple molecules and ions and identifying which ones have a non-planar structure.
Concepts Tested
📚 NCERT Sections This Tests
3.23 — The Rate Constant For The Decomposition Of Hydrocarbons Is 2.418 × 10–5S–1
Chemistry Class 11 · Chapter 3
3.23 The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5s–1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
5.18 — What Is Crystal Field Splitting Energy? How Does The Magnitude Of Do Decide
Chemistry Class 11 · Chapter 5
5.18 What is crystal field splitting energy? How does the magnitude of Do decide the actual configuration of d orbitals in a coordination entity?
1.34 — Vapour Pressure Of Water At 293 K Is 17.535 Mm Hg. Calculate The Vapour
Chemistry Class 11 · Chapter 1
1.34 Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
📋 Question Details
- Chapter
- Chemical Bonding
- Topic
- Molecular geometry and VSEPR theory
- Year
- 2022
- Shift
- 27 Jul Shift 2
- Q Number
- Q53
- Type
- Numerical
- NCERT Ref
- Class 11 Chemistry Ch 4: Chemical Bonding and Molecular Structure
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