`1.0 g` of `Mg "atom" ("atomic mass" = 24.0 "amu")` in the vapour phase absorbs `50.0 kJ` energy .Find the composition of the final maximum if the first and the second if of Mg are `740 kJ "mol^(-1)` respectively

`1.0 g` of `Mg "atom" ("atomic mass" = 24.0 "amu")` in the vapour phas

1.	`1.0 g` of `Mg "atom" ("atomic mass" = 24.0 "amu")` in the vapour phase absorbs `50.0 kJ` energy .Find the composition of the final maximum if the first and the second if of Mg are `740 kJ "mol^(-1)` respectively
Answer» Mole of Mg `= 1//24` `= 4.167 xx 10^(-2) mol` Energy used in ionising Mg to `Mg^(o+)` `= (4.167 xx 10^(-2)) xx (740 kJ mol ^(-1))` `= 30.84 kJ` Energy used in ionising `Mg^(o+)` to `Mg^(2+)= (50.0 - 30.84 )kJ =- 19.16 kJ` Amount of `Mg^(o+)`converted to `Mg^(2+)` `= (19.16 xx kJ)/(1450 kJ mol^(-1)) 1.321 xx 10^(-2) mol` Amount of `Mg^(o+)` remaining as each `(4.167 xx 10^(-2) - 1.327 xx 10^(-2)) mol = 2.846 xx 10^(-2) mol` % of `Mg^(o+) = (2.846 xx 10^(-2))/(4.167 xx 10^(-2)) xx 100 = 68.3` % of `Mg^(2+) = 100 - 68.3 = 31.7`

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`1.0 g` of `Mg "atom" ("atomic mass" = 24.0 "amu")` in the vapour phase absorbs `50.0 kJ` energy .Find the composition of the final maximum if the first and the second if of Mg are `740 kJ "mol^(-1)` respectively

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