One mole of magnesium in the vapor state absored `1200 kJ mol^(-1)` of enegry. If the first and second ionization energies of `Mg` are `750` and `1450 kJ mol^(-1)`, respectively, the final composition of the mixture is

Options

A. `69% Mg^(+), 31% Mg^(2+)`
B. `59% Mg^(+), 41% Mg^(2+)`
C. `49%Mg^(+), 51% Mg^(2+)`
D. `29% Mg^(+), 71% Mg^(2+)`

Answer

A