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 1g of Mg = 1/24 = 0.0417
1g of Mg (g) absorbs `=(1200)/(24)= 50kJ`
`therefore` Energy required to convert Mg (g) to `Mg^(+)` (g) `= 0.0417 xx 750 = 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) xx 100 = 68.82%`
`% Mg^(2+) = 100- 68.82 = 31.18%`