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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» <html><body><p></p>Solution :No. of moles of Mg `=( "Mass")/( "<a href="https://interviewquestions.tuteehub.com/tag/atomic-2477" style="font-weight:bold;" target="_blank" title="Click to know more about ATOMIC">ATOMIC</a> mass") = (1.2)/( 24) = 0.05` <br/>Suppose final mixture contains 'a' moles of `Mg^(+)` ions and 'b' moles of `Mg^(2+)` ions. Then <br/> `a+b=0.05`….(i)<br/>`Mararr Mg^(+) ` <a href="https://interviewquestions.tuteehub.com/tag/absorb-846161" style="font-weight:bold;" target="_blank" title="Click to know more about ABSORB">ABSORB</a> `<a href="https://interviewquestions.tuteehub.com/tag/750-335306" style="font-weight:bold;" target="_blank" title="Click to know more about 750">750</a> kJ mol^(-1)` and `Mg rarr Mg^(2+)`absorb `750 + 1450 = 2200 kJ mol^(-1)`<br/> Hence, `a xx 750 + b xx 2200 = <a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a>` ( Given ) <br/> or ` 15 a + 44b = 1`....(ii) <br/> or ` 15 a + 44 (0.05 -a) = 1` <br/> or ` 29a = 1.2 ` or `a= 0.04`<br/>`:.b = 0.05 - 0.04 = 0.01` <br/> `%`mole of `Mg^(+). ( 0.04)/( 0.05) xx 100 = 80%, % ` mole of `Mg^(2+) = 20%`</body></html>


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