__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 10^{23}) 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 6.023x10^{23}.

Therefore, one mole of any substance = 6.023 x 10^{23} particles may
be atoms, molecules, ions

For example: One
mole of oxygen atoms represents 6.023 x 10^{23} atoms of oxygen and 5
moles of oxygen atoms contain 5 x 6.023x10^{23} 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 CO_{2}

Solution: gram molecular mass of CO_{2}
= 44g

Number of molecules = (6.023 x 10^{23}
x 11) / 44 = 1.51 x 10^{23}
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 10^{23}

Problem: Calculate the mass of 18.069 x 10^{23}
molecules of SO_{2}

Solution: Gram molecular mass SO_{2}
= 64gm

The mass of 18.069 x 10^{23}
molecules of SO_{2} = (64x18.069 x 10^{23})/ (6.023 x 10^{23})
= 192 g

Do yourself: (a) Calculate
the mass of glucose in 2 x 10^{24} 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 10^{23}
molecules in it.

Solution: Number of moles = [Number of
molecules/(6.023 x 10^{23})] = (
3.0115 x 10^{23})/( 3.0115 x 10^{23}) =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 Al_{2 }O_{3} = 2x27 + 3x16 = 102gm

The number of ions present in 102 gm of
aluminum oxide = 6.023 x 10^{23}
ion

The number of ions present in 0.051g of
aluminum oxide= (6.023 x 10^{23} ion x 0.051g)/ 102 gm

=
6.023 x 10^{23} ion x0.0005 = 3.0115 x 10^{20} ions

In Al_{2 }O_{3}, Aluminium
and oxygen are in ratio 2:3

So, The number of aluminum ions present(Al^{3+})
in 0.051g of aluminum oxide = 2 x 3.0115 x 10^{20} ions =6.023 x 10^{20}
ion

**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 x10^{22} molecules of CH_{4
}(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 10^{23} molecules

Therefore, 1 mole (6.02 x 10^{23} molecules)
of CH_{4} 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 10^{21 }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 10^{23} particles
of a substance in one mole of that substance. Thus

volume occupied by 6.022 X 10^{23} molecules
of oxygen = 22.4 L

volume occupied by 10^{21} molecule
of oxygen = [(22.4 X 10^{21}) / (6.022 X 10^{23})]

= 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:

2H_{2}O → 2H_{2} + O_{2}

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 H_{2}O
= 1x2 + 16 = 18 gm

18 gm of water contain 6.022 x 10^{23} molecules

1 gm of water will contain = (6.022 x 10^{23})/18 molecules
= 0.33 x 10^{23 }molecules

So, the no. of molecules of water in 1ml of
water = 3.3 x 10^{22}

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