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 9: Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g ....
Comments
Or practise isi type k numerical ki batana Nice method well
Answer is 23.1
Thanks uhhh
Ye answer book me nhi hai
Relative lowering kaise aaya
This is not the answer of my questions plz coelrrect it
Rlvp=p°1-p1/p°1= 23.8 -23.4/23.8= 0.4/23.8=0.0168 Rlvp =0.017(approx)
But the answer is 289.5 bar. Which is nentioned in 12th NCEET book.
Relative lowering kaise nikala vo bhi toh btao
vapour pressure in mm hg of 0.1 mole of urea in 180 g of water at 25 C is(the vapour pressure of water at 25 C is 24 mm hg)