1.2g of Mg atoms in vapour phase absorb `50.0 kJ` of energy to convert Mg atoms into Mg ions. The energy absorbed is needed for the following changes `:`
`Mg(g) rarr Mg^(+) (g) + e^(-) , DeltaH =750 kJ mol^(-1)`
`Mg^(+)(g) rarrMg^(2+) (g) + e^(-) , Delta H = 1450 kJ mol^(-1)`
Calculate the percentage of `Mg^(+)` and `Mg^(2+)` ions in the final mixture.

Explanation

No. of moles of Mg `=( "Mass")/( "Atomic mass") = (1.2)/( 24) = 0.05`
Suppose final mixture contains 'a' moles of `Mg^(+)` ions and 'b' moles of `Mg^(2+)` ions. Then
`a+b=0.05` ….(i)
`Mararr Mg^(+) ` absorb `750 kJ mol^(-1)` and `Mg rarr Mg^(2+)`absorb `750 + 1450 = 2200 kJ mol^(-1)`
Hence, `a xx 750 + b xx 2200 = 50` ( Given )
or ` 15 a + 44b = 1` ....(ii)
or ` 15 a + 44 (0.05 -a) = 1`
or ` 29a = 1.2 ` or `a= 0.04`
`:. b = 0.05 - 0.04 = 0.01`
`%` mole of `Mg^(+). ( 0.04)/( 0.05) xx 100 = 80%, % ` mole of `Mg^(2+) = 20%`