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. `13% Mg^(+) 87%Mg^(2+)`
B. `14%Mg^(+) +86% Mg^(2+)`
C. `69% Mg^(+) + 31% Mg^(2+)`
D. `31%Mg^(+) + 69% Mg^(2+)`

Answer

C

Explanation

`Mg rarrMg^(+) +e^(-), Delta_(i)H_(1) = 750 kJ mol^(-1)`. Thus, the enegry left after the ionization of `1 mol` of `Mg` vapor is
`1200 - 750 = 450 kJ`
Since `1450 kJ` causes the ionization of `1mol` of `Mg^(+)` to `Mg^(2+), 450kJ` will causes the ionization of
`(1)/(1450) xx 450 = 0.31 mol of Mg^(+)`
i.e., out of `1 mol` of `Mg^(+)`, only `0.31 mol` of `Mg^(+)` will be converted into `Mg^(2+)` leaving behind `0.69 mol` of `Mg^(+)`.