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.

1.2g of Mg atoms in vapour phase absorb 50.0 kJ of energy to convert M

1.	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.
Answer» Solution :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%`