One moleof magnesiumin thevapourstateabsorbed1200 kJ mol^(-1)of energy. It thefirstand secondionizationenergiesof Mg are750and1450 kJ mol^(-1) respectively, the finalcompositionof themixture is

One moleof magnesiumin thevapourstateabsorbed1200 kJ mol^(-1)of energy

1.	One moleof magnesiumin thevapourstateabsorbed1200 kJ mol^(-1)of energy. It thefirstand secondionizationenergiesof Mg are750and1450 kJ mol^(-1) respectively, the finalcompositionof themixture is
Answer» `31%MG^(+)+ 69 % Mg^(2+)` `69% Mg^(+)+ 31 %Mg^(2+)` `86 % Mg^(+) + 14% Mg^(2+)` `14% Mg^(+)+ 86%Mg^(2+)` SOLUTION :Energyabsorbed inthe ionizationof 1 moleof Mg (g)to `Mg^(+)` ( g) =750 k J Energyleft UNUSED= 1200- 750kJ `% ` of`Mg^(+) (g)`convertedinto `Mg^(2+)(g)` `= (450 )/(1450 )xx 100= 31%` Thusthe% of`Mg^(+)(g) =100 - 31 = 69 %`