First and second ionisation energies of `Mg (g)` are 740 and `1450kJ mol ^(-1).` Calculate percentage of `Mg^(+)(g)and Mg ^(2+) (g), ` if 1 g of Mg(g) absorbs 50 kJ of energy.

Explanation

Number of moles of 1g of `Mg = 1/24 = 0.0417`
Energy required to convert mg(g) to `Mg ^(+)(g) =0.0417 x 740 =30.83kJ`
Remaining energy `= 50 -30.83 =19.17 kJ`
Number of moles of `Mg ^(2+)` formed `= (19.17)/(1450) =0.0132`
Thus, remaining `Mg ^(+)` will be = `0.0417 - 0.0132=0.0285`
`%Mg ^(+) = (0.0285)/(0.0417) xx 100= 68.35%`
`%Mg ^(+2) =100 -68.35 =31. 65%`