`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)`