How many mL of 0.1 M HCl are required to | 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

  • Rishabh
  • 2020-03-19 09:28:20

Super duper smarttt


  • Ravi
  • 2019-08-05 22:59:05

Easy understanding method nice. Diaper for difficult questions..


  • Krysta
  • 2019-05-30 07:42:44

Perfect 👍


  • Kunalsharma
  • 2019-05-03 16:45:08

Good but it can be done in short also😍


  • Uttam
  • 2019-04-21 21:11:30

Thanks


  • Uttam
  • 2019-04-21 21:11:10

Thanks


  • Vikas
  • 2019-04-20 23:37:02

Thanks


  • Girjesh kumar
  • 2019-04-11 08:29:19

Very nice.


  • KRISHAN KUMAR
  • 2019-04-11 05:34:49

Hmm thank you. But still.........


  • Facebook
  • 2019-03-30 18:12:53

You are help me to solve this prnblem so thank you for it


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 6: How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 cont....