Calculate the number of aluminium ions present in 0.051 g of aluminium oxide

Explanation

Molecular mass of aluminium oxide `(Al_(2) O_(3))`
`= 2 xx 27 + 3 xx 16`
`=54 + 48`
=102 u
Molar mass (M) of `Al_(2) O_(3) = 102` g
Given mass (m) of `Al_(2) O_(3) = 0.051` g
Avogadro.s number `(N_(0)) = 6.023 xx 10^(23)`
Number (N) of particles (ions) = `("Given mass ")/("Molar mass") xx "Avogadro.s number"`
`implies N = (m)/(M) xx N_(0)`
`implies N = (0.051)/(102) xx 6.023 xx 10^(23)`
`implies N = 0.0030 xx 10^(23)`
Since each molecules of `Al_(2) O_(3)` contains two aluminium ions , so number of aluminium ions
`= 2 xx 0.0030 xx 10^(23)`
`= 0.006 xx 10^(23)`
`= 6 xx 10^(20)`