| ⇦ |
| ⇨ |
Volume occupied by one molecule of water (density= 1 g cm⁻³) is
Options
(a) 9 x 10⁻²³ cm³
(b) 6.023 x 10⁻²³ cm³
(c) 3 x 10⁻²³ cm³
(d) 5.5 x 10⁻²³ cm³
Correct Answer:
3 x 10⁻²³ cm³
Explanation:
density = mass/volume
volume=mass/density
=1 gram/ 1 gram cm⁻³ = 1 cm³
1 g water= 1/18 moles of water
(at wt of H *2 + at wt of oxygen = 18)
volume occupied by 1 molecules of water = (18 / (6.02 x 10²³)) cm³
= 3 x 10⁻²³ cm³
Related Questions: - Complexes with bidentate ligands are called
- The most effective electrolyte for the coagulation of As₂S₃ sol is
- The weight of one molecule of a compound C₆₀H₁₂₂ is
- The rate of reactions exhibiting negative activation energy
- Who modified Bohr’s theory by introducing elliptical orbits for electron path
Question Type: Memory
(964)
Difficulty Level: Easy
(1008)
Topics: Basic Concepts of Chemistry
(94)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Complexes with bidentate ligands are called
- The most effective electrolyte for the coagulation of As₂S₃ sol is
- The weight of one molecule of a compound C₆₀H₁₂₂ is
- The rate of reactions exhibiting negative activation energy
- Who modified Bohr’s theory by introducing elliptical orbits for electron path
Question Type: Memory (964)
Difficulty Level: Easy (1008)
Topics: Basic Concepts of Chemistry (94)
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