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
