## 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)

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)
3. NCERT Questions

NCERT Questions

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