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)

Explanation

Step 1: Calcilation of no. of moles in 0.0591g of `Al_(2)O_(3)`
Gram molecular mass of `Al_(2)O_(3)=2xx"Grams atomic mass of "Al+3xx"Gram atomic mass of O"`
`=(2xx27g)+(3xx16g)=102g`
Now,a 102g of `Al_(2)O_(3)` =1 mol
`therefore 0.051g of Al_(2)O_(3)=((0.051g))/((102g))xx(1"mol")=0.0005"mol"`
Step -II: Calculation of no. of Al ions in 0.001 mole of `Al_(2)O_(3)`
1 mole of `Al_(2)O_(3)` contains Al atoms `=2xxN_(0)`
`0.005"mole of"Al_(2)O_(3) "contains Al atoms"=2xx0.005xxN_(0)`
`=2xx0.005xx6.022xx10^(23)=6.022xx10^(20)"atoms"`
In `Al_(2)O_(3)` valency of Al=3+
No. of aluminium ions (`Al^(3+))` present in the same as the no. of Al atoms.