Ist and IInd IE Of Mg Are 7.646 And 15.035 EV Respectively. The Amount Of Energy Needed To Convert All The Atoms Of Magnesium Into Mg2+ ions Present In 12 Mg Of Magnesium Vapours Is [Given, 1 EV = 96.5 KJ Mol–1]

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Question

Ist and IInd IE of Mg are 7.646 and 15.035 eV respectively. The amount of energy needed to convert all the atoms of magnesium into Mg2+ ions present in 12 mg of magnesium vapours is [Given, 1 eV = 96.5 kJ mol–1]

Solution

Correct option is

1.1

 

Total energy required for the conversion of one Mg atom into Mg2+ is = IE1 + IE2

             = 7.646 + 15.035 eV   

             = 22.681 eV   

             = 2188.6 kJ mol–1

  

                         = 0.5 × 10–3   

∴ The energy required to convert 0.5 × 10–3 mol Mg into

Mg2+ = 0.5 × 10–3 × 2188.6

         = 1.09 ≈ 1.1.

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