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

Answer

B

Explanation

Number of moles of 1 g of Mg=1/24=0.0417
1 g of Mg (s) absorbs = `(1200)/(24)=50 kJ`
`therefore` Energy required to convert Mg (g) to `Mg^(+)`(g)
`=0.0417xx750=31.275 kJ`.
`therefore` Remaining energy = 50 - 31.275 = 18.725 kJ
Number of moles of `Mg^(2+)` formed = `(18.725)/(1450)=0.013`
Thus remaining `Mg^(+)` will be = `0.0417 - 0.013 = 0.0287`
`therefore % Mg^(+)=(0.0287)/(0.0417)xx100=68.82%`
`% Mg^(2+)=100-68.82 = 31.18%`.