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  • The ionization enthalpies of Na & K are 495KJ/mol & 419KJ/mol respectively. The energy required to convert all the atoms present in 23 mg of Na vapours & 39 mg of K vapours of their res

The ionization enthalpies of Na & K are 495KJ/mol & 419KJ/mol respectively. The energy required to convert all the atoms present in 23 mg of Na vapours & 39 mg of K vapours of their respective gaseous cations respectively are:

Option: 1

495J & 419J


Option: 2

4.95J & 4.19J


Option: 3

495KJ & 419KJ


Option: 4

49.5J & 41.9J


Answers (1)

No. of moles of Na =\frac{23}{100*23}=10^{-3}

No. of moles of K =\frac{39}{100*39}=10^{-3}

\therefore the amounts of energies required for 10^{-3} mole each of Na & K are 495*10^{-3}KJmol^{-1}\: \: and\: \: 419*10^{-3}KJmol^{-1}  or 495J & 419J respectively.

Posted by

Ramraj Saini

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