Calculate the mass of sodium acetate (CH | Class 11 Chemistry Chapter Some Basic Concepts of Chemistry, Some Basic Concepts of Chemistry 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

  • Ridham Patel
  • 2020-05-29 18:45:51

Thank you


  • Tanu
  • 2020-05-24 19:30:46

Thanks a lot


  • Adi
  • 2019-07-21 07:53:18

Thanks for elaboration


  • Indranil
  • 2019-07-09 15:09:38

Thanks


  • Saumya
  • 2019-07-07 13:21:56

thanks for such good explanation


  • Prasoon
  • 2019-06-30 14:40:45

Supriti try to love


  • ravi
  • 2019-06-23 13:44:59

thnx


  • ravi
  • 2019-06-23 13:44:11

thanx


  • Sahil
  • 2019-06-18 17:24:38

Why we multi. 82.0245 with 0.1875 not divede


  • sakshi
  • 2019-06-05 09:25:20

Thank you but I want it a little more briefly


Comment(s) on this Question

Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 5: Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solut....