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The IE1 " and " IE2 of Mg (g) are 740 a...

The `IE_1 " and " IE_2` of Mg (g) are 740 and `1450 kJ "mol:^(-1)` . Calculate the percentage of `Mg^(+)` (g) and `Mg^(2+)` (g) if 1g of Mg (g) absorbs 50 kJ of energy.

Text Solution

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Number of moles of 1 g of Mg = `(1)/(24) = 0.0417`
Energy required to convert Mg (g) to `Mg^(+) (g) = 0.0417 xx 740 = 30.83` kJ
Remaining energy = `50 - 30.83 = 19.17` kJ
Number of moles of `Mg^(2+)` formed = `(19.17)/(1450) = 0.0132`
Thus , remaining `Mg^(+)` will be = `0.0417 - 0.0132 = 0.0285`
% `Mg^(+) = (0.0285)/(0.0417) xx 100 = 68.35%`
% `Mg^(+2) = 100 - 68.35 = 31.65 %`
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