Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
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
mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16
= 102 g
i.e., 102 g of Al2O3= 6.022 × 1023molecules of Al2O3
Then, 0.051 g of Al2O3contains = 6.022 × 1023 / 102 × 0.051 molecules
= 3.011 × 1020 molecules of Al2O3
The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.
Therefore, the number of aluminium ions (Al3+) present in 3.011 × 1020molecules (0.051 g ) of aluminium oxide (Al2O3) = 2 × 3.011 × 1020
= 6.022 × 1020
