`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 magnesium, if the first and the second IE of Mg are `740 kJ mol^(-1)` and `1450 kJ mol^(-1)`respectively.

Answer

67

Explanation

`1g = (1)/(24)` moles

`x xx 720 +((1)/(24)-x) (720 + 1440) = 50`
`720 x +(2160)/(24) - 2160x = 50`
`rArr 1440 x = 90 - 50 = 40`
`rArr x = (40)/(1440) = (1)/(36)`