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

`6.022 xx 10^(20)` ions

Explanation

mole of aluminium oxide `(Al_(2)O_(3))` = 2 × 27 + 3 × 16 = 102g
i.e., 102g of `Al_(2)O_(3) = 6.022 × 10^(23)` molecules of `Al_(2)O_(3)`
Then, 0.051 g of `Al_(2)O_(3)` contains `=6.022xx10^(23)//102xx0.051` molecules
`= 3.011 × 10^(20)` molecules of `Al_(2)O_(3)`
The number of aluminium ions `(Al^(3+))` present in one molecules of aluminium oxide is 2.
Therefore, The number of aluminium ions `(Al^(3+))` present in `3.11 × 10^(20)` molecules (0.051g) of aluminium oxide `(Al_(2)O_(3)) = 2 × 3.011 × 10^(20)`
`= 6.022 × 10^(20)`