Vapour pressure of water at 293 Kis 17.5 | Class 12 Chemistry Chapter Solutions, Solutions NCERT Solutions

Question:

Calculate the molecular mass of the following:
(i) H2O
(ii) CO2
(iii) CH4

Answer:

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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Comments

  • anonymous
  • 2021-04-10 14:50:18

very good


  • Abhi
  • 2019-06-16 11:29:11

Did\'nt understand the last second step how does 0.097 comes? Please answer....


  • Aqeel Ahmad
  • 2019-05-30 00:52:25

Bole to jhakkas


  • shamim abbas laskar
  • 2019-02-13 19:23:39

thnx


  • Rashmika
  • 2018-10-25 19:49:23

Irrelevant answer .. Waste of time... Not at all satisfied...


Comment(s) on this Question

Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 34: Vapour pressure of water at 293 Kis 17.535 mm Hg. Calculate the vapour pressure of water at 293 Kwhe....