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Calculate the percentage of Mg((g))^(o+)...

Calculate the percentage of `Mg_((g))^(o+)` and `Mg_((g))^(2+)` if `2.4 g` of `Mg` absorbs `120 kJ` of energy. The `IE_(1)` and `IE_(2)` of `Mg_((g))` are `740` and `1450 kJ mol^(-1)`.

Text Solution

Verified by Experts

Moles of `Mg = (2.4)/(24) = 0.1`
Energy required in the conversion of `0.1` mol of `Mg_((g))` to `Mg_((g))^(o+)`.
`= 740 xx 0.1 = 74 kJ`
Energy left unused `= 120 - 74 = 46 kJ`
So, `46 kJ` of energy will be used to ionise `Mg_((g))^(o+)` to `Mg_((g))^(2+)`
`:.` Number of moles of `Mg_((g))^(o+)` converted into
`Mg_((g))^(2+) = (46)/(1450) = 0.03`
Number of moles of `Mg_((g))^(o+)` left `= 0.1 - 0.03 = 0.07`
`:. % of Mg_((g))^(o+) = (0.07)/(0.1) xx 100 = 70%`
and `% of Mg_((g))^(2+) = 100 - 70 = 30%`
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