`1g` of magnesium vapor absorbs `50 kJ` of light enegry. Find the percentage of `Mg^(+)(g)`and `Mg^(2+)(g)` in the vapor if `Delta_(i)H_(1) = 738kJ mol^(-1)`and `Delta_(i)H_(2) = 1450kJmol^(-1)`.
Strategy: Find the number of moles of `Mg` atoms present in `1g` of `Mg` vapor. Find the moles of `Mg^(+)` and `Mg^(2+)` formed by absorbing enegry. Use the following ratio of calculate the percentage of `Mg^(+)(g)`:
`%Mg^(+)(g) = (n_(Mg^(+(g))))/(n_(Mg^((g))))xx100%`
Finally, `% Mg^(2+) = 100% - % Mg^(+)(g)`

Explanation

Step `1: n_(Mg) = (Mass_(Mg))/("Molar mass"_(Mg)) = (1g)/(24g mol^(-1)) = 0.0417 mol`
Step `2:` Energy required to ionize `Mg` to `Mg^(+)`
`= (n_(Mg))(Delta_(i)H_(1))`
`= (0.0417 mol)(738 kJ mol^(-1))`
`= 30.8 kJ`
Step `3:` Energy left for `2nd` ionization `= 50 kJ - 30.8 kJ = 19.2 kJ`
Step `4`: Moles of `Mg^(+)` ionized to `Mg^(2+)`
`= ("Enegry left")/(Delta_(i)H_(2)) = (19.2kJ)/(1450kJ mol^(-1))`
`= 0.0132 mol`
Step `5`: Moles of `Mg` present as `Mg^(+) = n_(Mg) - n_(Mg^(2+))`
`= (0.0417) - (0.0132)`
`= 0.0285`
Step `6: % of Mg^(+)(g) = (n_(Mg^(+)))/(n_(Mg))= (0.0285)/(0.417) xx 100%`
`= 68.35%`
Step `7: % of Mg^(2+)(g) = (100%) - (%Mg^(+))`
`= (100%) - (68.35%)`
`= 31.65%`