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One mole of magnesium in the vapor state...

One mole of magnesium in the vapor state absorbed `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

A

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

B

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

C

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

D

`13% Mg^(+) 87% Mg^(2+)`

Text Solution

Verified by Experts

The correct Answer is:
B
Energy absorbed in the ionisation of 1 mole of Mg(g) to `Mg^(+)` (g) =750 k J
Energy left unused =1200-750 =45 0 kJ
% of `Mg^(+)` (g) converted into `Mg^(2+)`(g)
`=450/1450xx100=31%`
thus the percentage of `Mg(+)(g)`
=100-31=69 %
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