| ⇦ | 
| ⇨ |
0.037g of an alcohol, R-OH was added to C₂H₅Mgl and the has evolved measured 11.2cc at STP. The molecular mass of R-OH will be
Options
(a) 47
(b) 79
(c) 77
(d) 74
Correct Answer:
74
Explanation:
R—OH + C₂H₅Mgl → C₂H₆ + RMgl
0.037 11.2 cc
11.2 cc gas evolved by alcohol= 0.037 g
22400 cc gas will be evolved by alcohol
= 0.037/11.2 x 22400
= 74 g
Related Questions: - The oxyacid of phosphorus, in which phosphorus has the lowest oxidation state, is
- Ethyl bromide can be obtained by the action of HBr on
- The constituent of light oil is
- This radical can be identified by Borax bead test is
- ³⁵₁₇Cl and ³⁷₁₇Cl are two isotopes of chlorine. If average atomic weight is 35.5
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
- The oxyacid of phosphorus, in which phosphorus has the lowest oxidation state, is
- Ethyl bromide can be obtained by the action of HBr on
- The constituent of light oil is
- This radical can be identified by Borax bead test is
- ³⁵₁₇Cl and ³⁷₁₇Cl are two isotopes of chlorine. If average atomic weight is 35.5
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