`1.0 g` of `Mg "atom" ("atomic mass" = 24.0 "amu")` in the vapour phase absorbs `50.0 kJ` energy .Find the composition of the final maximum if the first and the second if of Mg are `740 kJ "mol^(-1)` respectively

Explanation

Mole of Mg `= 1//24`
`= 4.167 xx 10^(-2) mol`
Energy used in ionising Mg to `Mg^(o+)`
`= (4.167 xx 10^(-2)) xx (740 kJ mol ^(-1))`
`= 30.84 kJ`
Energy used in ionising `Mg^(o+)` to `Mg^(2+)= (50.0 - 30.84 )kJ =- 19.16 kJ`
Amount of `Mg^(o+)`converted to `Mg^(2+)`
`= (19.16 xx kJ)/(1450 kJ mol^(-1)) 1.321 xx 10^(-2) mol`
Amount of `Mg^(o+)` remaining as each
`(4.167 xx 10^(-2) - 1.327 xx 10^(-2)) mol = 2.846 xx 10^(-2) mol`
% of `Mg^(o+) = (2.846 xx 10^(-2))/(4.167 xx 10^(-2)) xx 100 = 68.3`
% of `Mg^(2+) = 100 - 68.3 = 31.7`