1.

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

Answer»

`31% Mg^(+)+69% Mg^(2+)`
`69%Mg^(+)+31% Mg^(2+)`
`86%Mg^(+)+14% Mg^(2+)`
`14%Mg^(+)+86%Mg^(2+)`

Solution :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%`.


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