Determine the enegry nedded to convert all the atoms of aluminium to aluminium tripostovity ions present in `27 mg` of aluminium vapors using the data: `Delta_(i)H_(1) = 577kJ mol^(-1), Delta_(i)H_(2) = 1820kJmol^(-1), Delta_(i)H_(3) = 2750kJ mol^(-1)`.
Strategy: Find the number of moelus of `A1` atoms present in `27 mg` of `A1` vapor and multiply it with enegry required to convert `1mol` of `A1` atoms into `1mol` of `A1^(3+)` ions. The enegry required is the sun of first, second, and third ionization enthalpies.

Explanation

Step 1:
`{:(Al(g)toAl^(+)(g)+e^(-)(g),,Delta_(i)H_(1)=577kJmol^(-)),(Al^(+)(g)toAl^(+)(g)+e^(-)(g),,Delta_(i)H_(2)=1820kJmol^(-)),(Al^(2+)(g)toAl^(3+)(g)+e^(-)(g),,Delta_(i)H_(3)=2750kJ mol^(-1)),(bar(Al(g)toAl^(3+)(g)+e^(-)(g)),bar(Delta_(i)H_(t)=5147kJmol^(-1))):}`
Step `2: n_(A1) = (Mass_(A1))/("Molar mass" _(A1)) = (27xx10^(-3)g)/(27g mol^(-1)) = 10^(-3)mol`
Step `3:` Amount of enegry required
`=(n_(A1))`(Enegry required per mol)
`= (10^(-3)mol) (5147kJ mol^(-1))`
`= 5.147 kJ`