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Calculate the energy required to convert all atoms of `Mg` to `Mg^(2+)` ions present in `48 mg` of `Mg` vapours. `IE_(1)` and `IE_(2)` of `Mg` are `740` and `1450 kJ mol^(-1)` respectively.

Text Solution

Verified by Experts

`Mg_((g)) rarr Mg_((g))^(o+) + e^(-), 1E_(1) = 740 kJ mol^(-)`
`Mg_((g))^(+) rarr Mg_((g))^(2+) + e^(-), 1E_(2) = 1450 kJ mol^(-)`
`:.` Total energy required to convert `1` mol of `Mg_((g))` into
`Mg_((g))^(2+) ion = IE_(1) + IE_(2)`
`= (740 + 1450) kJ mol^(-)`
`= 2190 kJ mol^(-)`
`48 mg of Mg = (48 xx 10^(-3))/(24) mol`
`= 2 xx 10^(-3) mol`
`:.` Energy required to ionise `20 xx 10^(-3)` mol of `Mg_((g))`
`= 2190 xx 2 xx 10^(-3)`
`= 4.380 kJ`
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