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.

MORE TO KNOW

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 10²³x 11) / 44 = 1.51 x 10²³molecules

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 10²³molecules of SO₂

Solution: Gram molecular mass SO₂= 64gm

The mass of 18.069 x 10²³molecules 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 10²³molecules 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 10²³ion

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 10²⁰ion

MORE TO KNOW

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

Related post to 9th Atoms and Molecules
IX(9th)Chemistry Ch 03: Solved Questions9th Atoms and molecules Study Notes, Q & ACBSE Test paper-1CBSE Test paper-2CBSE Test paper-3Numerical based on mole conceptSolved questions For FA-3Self-evaluation on Mole conceptQuestion bank By CBSE Pathshala For CBSEGuess Questions