Calculate the number of aluminium ions present in 0.051 g of aluminium 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)

 

Answer

We know that,

Number of moles of a substance = Given mass of the substance / Molar mass of the substance

                                                               n  = m/M ----- (1)

Also,

Number of moles of a substance = Number of particles of the substance / Avogadro's Constant

                                                                n  = N / N O

n  = N / 6.022 x 10 23 ------- (2)

 

From the above 2 formulae, we can say that,

 n = m / M = N / 6.022 x 10 23   ---- (3)



Given mass of Aluminium ion (Al 2 O 3 )= m = 0.051g

Molar mass of Aluminium Oxide (Al 2 O 3 ) = M = 102g

 

We need to find Number of ions, ie, N

 

Putting values in (3), we get

0.051 / 102 = N / 6.022 x 10 23  

N = 6.022 x 10 23   x (0.051 / 102)

   = 6.022 x 10 23   x (51 x 10 -3 ) / 102

   = 3.011 x 10 (23-3) 

   = 3.011 x 10 20   molecules

 

1 molecule of Al 2 O 3 has 2 Al 3+ ions

So, number of Al 3+ ions in 0.051 molecules of Al 2 O 3 is;

 

3.011 x 10 20 x 2 = 6.022 x 10 20

 

Thus, 6.022 x 10 20 Al 3+ ions are present in 0.051 molecules of Al 2 O 3

  1. Class 9
  2. Chapter 3 Class 9 - Atoms And Molecules (Term 2)

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CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 11 years and a teacher from the past 11 years. He teaches Science, Accounts and English at Teachoo