(IE)1$$ and $$(IE)2$$ of $$Mg(g)are$$ $$740$$, $$1540$$ kJ $$mol-1$$. Calculate percentage of $$Mg+(g) and $$Mg2+(g)$$ if $$1$$ g of $$Mg(g)$$ absorbs $$50.0$$ kJ of energy.

Question

Infinity Learn · Accepted Answer

Energy supplied per gram  $$=$$  $$50KJEnergy$$ supplied per mole  $$=$$  $$(50x24)KJ$$                                           $$=$$  $$1200$$ $$KJIP1$$ of "Mg" is $$740$$ $$KJ/MoleThus$$ all the gaseous "Mg" atoms completely get ionized to gaseous $$Mg+$$ ions by absorbing $$740KJ$$ ;Of the $$1200$$ KJ  energy supplied,The energy still available $$=$$ $$(1200-740)KJ$$                                        $$=$$ $$460$$ KJSince $$IP2$$ of "Mg" is $$1540$$ $$KJ/mole$$,$$Mg+g$$→$$Mg+2g$$,$$100$$% conversion Occurs only when $$1540$$ KJ of energy is supplied.∵only $$460KJ$$ of energy is available, the % of conversion of $$Mg+g$$→$$Mg+2g$$ will $$be4601540$$×$$100=29.87$$%Percentage $$Mg+2g$$ $$=$$ $$29.87$$%Percentage $$Mg+g$$   $$=$$ $$70.13$$%

(IE)1 and (IE)2 of Mg(g)are 740, 1540 kJ mol-1. Calculate percentage of Mg+(g) and Mg2+(g) if 1 g of Mg(g) absorbs 50.0 kJ of energy.

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