| ⇦ | 
| ⇨ |
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: - Surface water contains
- The latest technique for the purification of organic concept is
- The number of π-bond present in propyne is
- When Cl₂ gas reacts with hot and concentration sodium hydroxide solution,
- If Ksp for HgSO₄ is 6.4 x 10⁻⁵, then solubility of the salt is
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
- Surface water contains
- The latest technique for the purification of organic concept is
- The number of π-bond present in propyne is
- When Cl₂ gas reacts with hot and concentration sodium hydroxide solution,
- If Ksp for HgSO₄ is 6.4 x 10⁻⁵, then solubility of the salt is
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