A mixture contains `F` and `Cl` atoms . The removal of an electron form each atom of the sample requires `28 kJ` while addition of an electron to each atom of mixture releases `68.8 kJ` energy .Calcualte the % composition of mixture .Given `IE` per atoms for `F` and `Cl` are `27.91 xx10^(-22)kJ` and `20.77 xx10^(-22) kJ`.
Electron gain enthaply for F and Cl are ` -5. 31 xx10 ^(-22) kJ` and `-5.78 xx 10^(22) kJ` respectivley

Explanation

Let the mixture contain x and y mole of F and CI
a. `," (x xx 6.023 xx 10^(23) mol^(-1)) (27.91 xx 10^(23) KJ) + (y + 6.023 xx 10^(23)) (20.77 xx 10^(22) KJ) = 284 KJ`…..(1)
b. `(x xx 6.023 xx 10^(23))(5.23 xx 10^(-23))(y xx 6.23 xx 10^(-23)) (5.78 xx 10^(-22)KJ) = 5.78 KJ`....(ii)
`:. x(1681) + y(1251) = 284`
x(333.1) + y(348.1) = 68.8`
Solving for x and y we get `x = 0.076,y = 0.125 `
% of `F = (0.076 xx 100)/((0.076 + 0.125)) = 37.81`
`% of CI = 100 - 37.81 = 62.19`