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The Delta(i)H1) and Delta(i) H(2) of Mg...

The `Delta_(i)H_1)` and `Delta_(i) H_(2)` of Mg (g ) are 740 and 1450 kJ `mol^(-1)` . Calculate the percentage of `Mg^(+)` ( g) and `Mg^(2+)` ( g) if 1 g of Mg ( g) absorbs 50 kj of energy .

Text Solution

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No . Of moes of Mg vapours present in 1 g = `1//24 = 0.0417`
Energy absorbed in the ionization of 0.0417 mole of Mg ( g) to `Mg^(2+) ( g) = 0.0417 xx 740 = 30. 83 kJ`
Energy left unused = 50 - 30. 83 = 19 . 17 kj
Now 19. 17 kj will be used to ionize `Mg^(+) ( g) ` to `Mg^(2+)` ( g)
`:. ` No . of moles of `Mg^(+) (g )` converted into `Mg^(2+) ( g) = 19. 17 // 1450 = 0.0132`

No. of moles of magnesium ions left as `Mg^(+) ( g) = 0.0417 -0.0417 -0.0132 = 0.0285`
`:. ` % age of `Mg^(+) (g) = (0.0285 // 0.0417 ) xx 100 =68.35 % ` and % age of `Mg^(2+) ( g) = 100 -68 . 35 = 31 .65 %`
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