NCERT Question 11 - Chapter 3 Class 9 - Atoms And Molecules

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