24 g of magnesium in the vapour state absorb 1200 kJ of energy. If IE_1 and IE_2 of magnesium are 750 and 1450 kJ mol^(-1) respectively, the final composition of mixture is

24 g of magnesium in the vapour state absorb 1200 kJ of energy. If IE_

1.	24 g of magnesium in the vapour state absorb 1200 kJ of energy. If IE_1 and IE_2 of magnesium are 750 and 1450 kJ mol^(-1) respectively, the final composition of mixture is
Answer» 0.69 MOL `MG^(+)` and 0.31 mol `Mg^(2+)` 0.59 mol `Mg^(+)` and 0.41 mol `Mg^(2+)` 0.49 mol `Mg^(+)` and 0.51 mol `Mg^(2+)` 0.29 mol `Mg^(+)` and 0.71 mol `Mg^(2+)` Solution :Molar MASS of Mg =24 g `mol^(-1)` `:.` Amount of Mg=1 mol THUS, 750 kJ of energy is used up to convert 1 mol of Mg to `Mg^(+)` Energy used to convert `Mg^(+)` to `Mg^(2+)` `=1200-750 =450 kJ` `:.` Amount of magnesium converted to `Mg^(2+)` `=(450 kJ)/(1450) =0.31 mol` `:. ` Amount of magnesium converted to `Mg^(+)` =1-0.31 =0.69 mol