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


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