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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

Verified by Experts

The correct Answer is:
`68.35`
No. of moles of Mg vapours present in 1g = /1 24 = .0 0417
Energy absorbed in the ionization of 0.0417 mole of Mg (g) to `Mg^(+)(g)= 0.0417 xx 740 = 30.83kJ`
Energy left unused `=50- 30.83= 19.17`kJ
Now, 19.17kJ will be used to ionize `Mg^(+) (g) " to " Mg^(2+)(g)`
`therefore` 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- 0.0132= 0.0285`
`therefore` % 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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