worksheet on mole concept

class-9

Mole is defined as the amount of substance that contains as many specifi ed 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
(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

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

Problem: Calculate the mass of 18.069 x 10² molecules of SO₂

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 ion³ 

Problem:  Calculate the number moles for a substance containing 3.0115 x 10²³ molecules in it.

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 

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.

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