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One mole of magnesium in the vapour stat...

One mole of magnesium in the vapour state absorbed 1200 kJ 'm o l^(-1)' of energy. If the first and second ionization energies of 'Mg' are 750 and '1450 kJ mol^(-1)' respectively, the final composition of the mixture is

A

`31%Mg^(+)+69%Mg^(2+)`

B

`69%Mg^(+)+31%Mg^(2+)`

C

`86%Mg^(+)+14%Mg^(2+)`

D

`14%Mg^(+)+86%Mg^(2+)`

Text Solution

Verified by Experts

The correct Answer is:
B
`Mg rarr Mg^(+), E=750kJ`
Remaining energy = 1200 - 750 = 450 kJ
Energy needed to convert 1 mol of `Mg^(+)` to `Mg^(2+)=1450`
Number of moles of `Mg^(2+)` produced
`" "=1/1450 times 450`
`" "=0.31`
`" "=31%`
Number of moles of `Mg^(+)` produced = 1 - 0.31
`" "=0.69`
`" "=60%`
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