first and second ionisation energies of mg are 740 and 1450 jmol-1 calculate percentage of mg+and mg2+ of 1g of mg absorbs 50 kj of energy

Answer 1

Please check your post. I think that's 740 kJ/mol and 1450 kJ/mol and not J/mol. I would look at it this way.
Convert 1 g Mg to mols = about 0.04 but you can be more exact t6han that.
0.04 x 740 = about 30 kJ
That leaves you 20 kJ of the total of 50 kJ.
How much will that 20 do for the 2nd ion at 1450 each? That will be about 20/1450 or about 0.01 mol.
Total ionized = 0.01+0.04 = 0.05
That is about 20% and 80% for 2nd and 1st respectively. Check my thinking.

Answer 2

This answer is incorrect

Answer 3

No. Of moles of mg present in 1g = 1/24 = 0.0417
Energy absorbed in the ionization of 0.0417 mole of Mg to Mg+ = 0.0417×740 = 30.83 kJ
Energy left unused = 50- 30.83 = 19.17kJ