(IE)1 and (IE)2 of Mg(g)are 740, 1540 kJ mol-1. Calculate percentage of Mg+(g) and Mg2+(g) if 1 g of Mg(g) absorbs 50.0 kJ of energy.

Energy supplied per gram  =  50KJEnergy supplied per mole  =  (50x24)KJ                                           =  1200 KJIP1 of "Mg" is 740 KJ/MoleThus all the gaseous "Mg" atoms completely get ionized to gaseous Mg+ ions by absorbing 740KJ ;Of the 1200 KJ  energy supplied,The energy still available = (1200-740)KJ                                        = 460 KJSince IP2 of "Mg" is 1540 KJ/mole,Mg+g→Mg+2g,100% conversion Occurs only when 1540 KJ of energy is supplied.∵only 460KJ of energy is available, the % of conversion of Mg+g→Mg+2g will be4601540×100=29.87%Percentage Mg+2g = 29.87%Percentage Mg+g   = 70.13%