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

Energy absorbed in the ionization of 1 mole of Mg (g) to `Mg^(+)` ( g) = 750 k J
Energy left unused = 1200 - 750 kJ
`% ` of `Mg^(+) (g)` converted into `Mg^(2+) (g)`
`= (450 )/(1450 ) xx 100 = 31%`
Thus the % of `Mg^(+) (g) = 100 - 31 = 69 %`