(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  =  50KJ
Energy supplied per mole  =  (50x24)KJ
                                           =  1200 KJ
${\mathrm{IP}}_1$ of "Mg" is 740 KJ/Mole
Thus all the gaseous "Mg" atoms completely get ionized to gaseous ${\mathrm{Mg}}^{+}$ ions by absorbing 740KJ ;

Of the 1200 KJ  energy supplied,
The energy still available = (1200-740)KJ
                                        = 460 KJ
Since ${\mathrm{IP}}_2$ of "Mg" is 1540 KJ/mole,
${\mathrm{Mg}}^{+}\left(\mathrm{g}\right)\to {\mathrm{Mg}}^{+2}\left(\mathrm{g}\right)$,100% conversion Occurs only when 1540 KJ of energy is supplied.
$\because$only 460KJ of energy is available, the % of conversion of ${\mathrm{Mg}}^{+}\left(\mathrm{g}\right)\to {\mathrm{Mg}}^{+2}\left(\mathrm{g}\right)$ will be

$\frac{460}{1540}\times 100=29.87\%$
Percentage ${\mathrm{Mg}}^{+2}\left(\mathrm{g}\right)$ = 29.87%
Percentage ${\mathrm{Mg}}^{+}\left(\mathrm{g}\right)$   = 70.13%