Saturday, October 19, 2013

9th(CBSE) Ch. Atoms and Molecules - Mole Concept and Related Problems

MOLE CONCEPT [CBSE Class IX Chemistry]
While performing a reaction, to know the number. of atoms (or) molecules involved, the concept of mole was introduced. The quantity of a substance is expressed in terms of mole.
Definition of mole:  Mole is defined as the amount of substance that contains as many specified elementary particles as the number of atoms in 12g of carbon-12 isotope.
One mole is also defined as the amount of substance which contains Avogadro number (6.023 x 1023) of particles.
Avogadro number: Number of atoms or molecules or ions present in one mole of a substance is called Avogadro number. Its value is 6.023x1023.
Therefore, one mole of any substance = 6.023 x 1023 particles may be atoms, molecules, ions
For example: One mole of oxygen atoms represents 6.023 x 1023 atoms of oxygen and 5 moles of oxygen atoms contain 5 x 6.023x1023 atoms of oxygen.
Questions based on mole concept:
1. When the mass of the substance is given:  Use this formula: Number of moles = given mass/ atomic mass
(a). Calculate the number of moles in  (i) 81g of aluminium ii) 4.6g sodium  (iii) 5.1g of Ammonia  (iv) 90g of water  (v) 2g of NaOH 
Solution: (i) Number of moles of aluminum = given mass of aluminium / atomic mass of aluminium = 81/27 = 3 moles of aluminum [Rest Question do yourself]
(b) Calculate the mass of 0.5 mole of iron
Solution: mass = atomic mass x number of moles = 55.9 x 0.5 = 27.95 g
Do yourself:  Find the mass of 2.5 mole of oxygen atoms [ Mass = molecular mass x number of moles]
2. Calculation of number of particles when the mass of the substance is given:
Number of particles =( Avogadro number x given mass)/gram molecular mass
Problem: Calculate the number. of molecules in 11g of CO2
Solution: gram molecular mass of CO2 = 44g
Number of molecules = (6.023 x 1023 x 11) / 44 = 1.51 x 1023 molecules
Do yourself: Calculate the number of molecules in 360g of glucose
3. Calculation of mass when number of particles of a substance is given:
Mass of a substance = (gram molecular mass x number of particles)/6.023 x 1023
Problem: Calculate the mass of 18.069 x 1023 molecules of SO2
Solution: Gram molecular mass SO2 = 64gm
The mass of 18.069 x 1023 molecules of SO2 = (64x18.069 x 1023)/ (6.023 x 1023) = 192 g
Do yourself: (a) Calculate the mass of glucose in 2 x 1024 molecules (b) Calculate the mass of 12.046 x 1023 molecules in CaO
4. Calculation of number of moles when you are given number of molecules:
Problem:  Calculate the number moles for a substance containing 3.0115 x 1023 molecules in it.
Solution: Number of moles = [Number of molecules/(6.023 x 1023)]  = ( 3.0115 x 1023)/( 3.0115 x 1023) =0.5 moles
Do yourself: (a) Calculate number of moles in 12.046x 1022 atoms of copper (b) Calculate the number of moles in 24.092 x 1022 molecules of water.
Problem: Calculate the number of aluminum ions present in 0.051 g of aluminum oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al=27 u)
Solution: Mass of the 1 mole of Al2 O3  = 2x27 + 3x16 = 102gm
The number of ions present in 102 gm of aluminum oxide  = 6.023 x 1023 ion
The number of ions present in 0.051g of aluminum oxide= (6.023 x 1023 ion x 0.051g)/ 102 gm 
=  6.023 x 1023 ion x0.0005 = 3.0115 x 1020 ions
In Al2 O3, Aluminium and oxygen are in ratio 2:3
So, The number of aluminum ions present(Al3+) in 0.051g of aluminum oxide = 2 x 3.0115 x 1020 ions =6.023 x 1020 ion
Volume occupied by one mole of any gas at STP is called molar volume. Its value is 22.4 litres 22.4 litres of any gas contains 6.023 x 1023 molecules.
Problem: Calculate the volume occupied at STP by :- (a) 64 gram of oxygen gas (b) 6.02 x1022 molecules of CH4 (c) 5 moles of nitrogen gas
Solution: (a) One mole of a gas occupies 22.4 L volume at STP
Now, number of moles in 64 g oxygen gas = 64/32 = 2
Therefore, volume occupied by 2 moles(64 g) of oxygen gas = 2 x 22.4 L = 44.8 L
(b) 1 mole = 6.02 x 1023 molecules
Therefore, 1 mole (6.02 x 1023 molecules) of CH4 gas occupies 22.4 L volume at STP.
(c) One mole of a gas occupies 22.4 L volume at STP
Therefore, volume occupied by 5 moles of nitrogen gas = 5 x 22.4 L = 112 L
Problem: Calculate the volume at STP occupied by 1021 molecules of Oxygen?
Solution: The molar volume that is the volume occupied by one mole of gas is 22.4 L. We know there are 6.022 X 1023 particles of a substance in one mole of that substance. Thus
volume occupied by 6.022 X 1023 molecules of oxygen = 22.4 L
volume occupied by 1021 molecule of oxygen = [(22.4 X 1021) / (6.022 X 1023)] 
                                                                              = 3.72 X 10-2 L = 37.2 ml
Q. Calculate the mass and volume of oxygen at STP, which will be evolved on electrolysis of 1 mole of(18 g) of water.
Answer: Electrolysis of water will break it down in its component as Hydrogen and Oxygen
Balanced chemical reaction:
2H2O → 2H2 + O2
From above equation, 2 mole of water will evolve 1 mole of oxygen gas upon electrolysis. Therefore 1 mole of water will produce 1/2 mole of Oxygen.
Mass of Oxygen evolved = number of moles of Oxygen evolved × Molecular wt. of Oxygen =1/2 × 32 = 16 g
At STP 1 mole of any gas will occupy 22.4l volume.
Volume of Oxygen evolved = No. of moles × 22.4l = 1/2 × 22.4l = 11.2l.
Q. How many molecules are present in 1 ml of water?
Answer: we know that density of water is 1gm/ml.
Hence, 1 gm water will = 1 ml water.
Now, we have molecular mass of water H2O = 1x2 + 16 = 18 gm
18 gm of water contain 6.022 x 1023 molecules
1 gm of water will contain = (6.022 x 1023)/18 molecules = 0.33 x 1023 molecules
So, the no. of molecules of water in 1ml of water = 3.3 x 1022


Kriti Srivastava said...

it help me too much

Sanjana Singavarapu said...

these numericals are very helpful.thank you

Subhashree Palo said...

these numerical help me too much..thank u.

Subhashree Palo said...

these numerical help me too much..thank u.

rajesh kumar said...

Thanks for your help 😊

Radha Sekarun said...


chitt patel said...

Nice blog helping me

kenneth ken said...

the above questions helped me. this too helped in doing more practice: Have problems understanding the mole concept: check this out

samarth nyamgoud said...

i was mad reading my textbook i dint understand it. this helped me in my exam very very much

Unknown said...

veryyyyyyyyyy helpfulllllllllll

Unknown said...

thank you thank you thank you thank you thank you thank you thank you

Apws school said...

Thanks for sharing as it is an excellent post would love to read your future post -for more knowledge .. Play School in Bangalore | International Schools in Bangalore

Unknown said...

Thanks u so much


Thank you excellent

Sumedha Bhatia said...
This comment has been removed by the author.
ccgghhjhi said...


ccgghhjhi said...

i spilled my coffee while reading this

Unknown said...

Very good questions easy to understand

Kavitha G S said...

Lub u

Post a Comment

Related Posts Plugin for WordPress, Blogger...