Avogadro’s number and the mole concept

 

   Matter has different types of particles, such as atoms, ions, molecules, etc. An element is made up of atoms.

A mole is used to represent the number of atoms in an element just as a crate is used to represent 30 eggs. Avogadro’s number is the number of units in a mole of any substance and it is equal to 6.02 x 10²³. Therefore, one mole of any substance is represented by 6.02x10²³. A diatomic element like chlorine is 6.02x10²³ molecules of chlorine. One mole of tetraoxosulphate (VI) oxide and one mole of methane each represents 6.02x10²³ molecules of these compounds. Similarly, sodium and calcium elements each also contain 6.02x10²³ atoms.

Example Question 1

 How many atoms are there in 10g of sodium? (Na = 23, Avogadro’s number = 6.02x10²³)

Since one mole is equal to 6.02x10²³ and one mole is equal to relative atomic mass expressed in grams.

    1mole = 6.02x10²³

    1mole = 23g

23g   = 6.02x10²³

10g = 6.02x10²³/23 x 10/1 = 2.62x10²³

Example Question 2

Calculate the mass in grams of 3.4x10²¹ atoms of copper

(Cu = 63.5, Avogadro’s number = 6.02x10²³)

6.02 x 10²³  = 63.5g

3.4x10²¹ =   63.5/6.02x10²³ x 3.4x10²¹ = 0.35g of Cu

Avogadro’s law and Molar Volume of gases

Avogadro’s law states that the equal volume of all gases at standard temperate and pressure contain the same molecules. According to this law, one mole of any gas at s.t.p. has a volume of 22.4dm²³ or 22400cm³.

So, one mole of oxygen gas has 22.4dm²³ at s.t.p.

One mole of hydrogen at s.t.p has 22.4dm²³

Example Question 1

Calculate the volume of carbon (IV) oxide that would be liberated if 20g of sodium trioxocarbonate (IV) dissolved in dilute hydrochloric acid at S.T.P

(Na =23, Cl = 35.5, C=12, O=16, H= 1, Molar volume =22.4dm³)

Na₂CO_3(s) + 2HCl_(aq) → 2NaCl_(aq) + H₂O_(l) + CO_2(g)

 

From the equation,

 106g of sodium trioxocarbonate IV = 22.4dm³ of Carbon IV oxide

20g of sodium trioxocarbonate IV   = 22.4/106 x 20  = 4.23dm³

 

Example Question 2

Calculate the volume of HCl gas that can be obtained at S.T.P from 12g of sodium chloride. (H=1,Na=23, Cl =35.5,, molar volume at S.T.P = 22.4dm³)

2NaCl_(s) + H₂SO₄ → Na₂SO_4(aq) + 2HCl_(g) 

Molar mass of NaCl = 58.5g/mol

From the equation,

2x 58.5 of NaCl→ 2x22.4 dm³ of HCl gas

117 of NaCl   = 44.8 dm³

12g  = 44.8/117 x 12/1   = 4.595dm³