The `IE_1 " and " IE_2` of Mg (g) are 740 and `1450 kJ "mol:^(-1)` . Calculate the percentage of `Mg^(+)` (g) and `Mg^(2+)` (g) if 1g of Mg (g) absorbs 50 kJ of energy.

Explanation

`Mg(g) to Mg^(+)(g) +e^(-) IE_(1)=750kJ mol^(-1)" ".....(i)"(given)"`
`Mg^(+) (g) to Mg^(2+) (g)+e^(-) IE_(2)=1450"kJ mol"^(-1)............(i)"given"`
Amount of energy needed for the conversion of one mole `Mg (g)to Mg^+ (g)= 750 kJ,` therefore, amount of energy left for the conversion of `Mg^(+)(g) to Mg^(2+) (g)+e^(-), i.e, 1200-750=450kJ`.
From eq. (ii), 1450 kJ energy is required to convert 1 mole of `Mg^+ (g) to Mg^(2+) (g) + e `
450 kJ energy is required to convert, `(450)/(1450)" mol "=0.31" mole of "Mg^+ (g)to Mg^(2+) (g) + e`
Thus, `Mg^(2+)(g)` = 0.31 moles and `Mg^+ (g)` = 1 -0.31 =0.69 moles. Therefore, `Mg^+ (g) : Mg^(2+) =0.69: 0.31`