Calculate the molecular mass of the following:
(i) H2O
(ii) CO2
(iii) CH4
Molecules are made up of atoms & are quite small, therefore the actual mass of a molecule cannot be detrmined.It is expressed as the relative mass.C-12 isotope is used to express the relative molecular masses of substances.Thus molecular mass of a substance may be defined as: the average relative mass of its molecule as compared to the mass of carbon atom taken as 12 amu.
(i) H2O:
The molecular mass of water, H2O
= (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen)
= [2(1.0084) + 1(16.00 u)]
= 2.016 u + 16.00 u
= 18.016
= 18.02 u
(ii) CO2:
The molecular mass of carbon dioxide, CO2
= (1 × Atomic mass of carbon) + (2 × Atomic mass of oxygen)
= [1(12.011 u) + 2 (16.00 u)]
= 12.011 u + 32.00 u
= 44.01 u
(iii) CH4:
The molecular mass of methane, CH4
= (1 × Atomic mass of carbon) + (4 × Atomic mass of hydrogen)
= [1(12.011 u) + 4 (1.008 u)]
= 12.011 u + 4.032 u
= 16.043 u
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Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 3: Write structures of different dihalogen derivatives of propane.....
Comments
Awesome questions (__abbhinav__)
Thank you.... It's not so hard th question is..
2,3 Di-bromopropane , 3,3-Dibromopropane and 1,3-Dibromopropane never possible
Why cant we add other halogens?
Can we write other halogens(cl,I)?
Why only Br in question halogen is mentioned
Fyn
Why not 2,3-dibromopropane and 3,1-dibromopropane
Why not 3,3 dibromopropane
Very nice.. Tks