Question

The total no of ions present in 111 g of CaCl2 is ?

Answer 1

The molecular weight of CaCl2 is 111g. So there's 1 mole cations and 2 mole anions. The total number of ions present is expressed as 3 mole.

Answer 2

Given: the mass of CaCl₂(m) = 111g

To Find: number of ions present in 111 g of CaCl₂

Solution:

Number of ions in 1 molecule of CaCl₂ = 1 + 2 = 3 ions (1Ca²⁺,2Cl²⁻)

To find the number of ions in 111 g of CaCl₂, first, find the number of molecules of CaCl₂(N) in 111g of CaCl₂.

number of moles(n) = m/M (M - Molecular mass = 111g)

Applying the above formula:

n = 111/111

n = 1 mole

Also, n = N/Nₐ (Nₐ = Avagadro number = 6.023 × 10²³)

N - number of molecules in 1 mole of CaCl₂ =  n × Nₐ

Applying the formula

N = n × Nₐ

N = 1 × 6.023 × 10²³

N = 6.023 × 10²³ molecules

3 ions present in 1 molecule of CaCl₂

∴Number of ions in 6.023 × 10²³ molecules of CaCl₂ = 6.023 × 10²³ × 3

                                                                                       = 18.1 × 10²³ ions   or  

                                                                                             3 moles of ions