Question

The ionization enthalpies of $Li$ and $Na$ are $520$ kJ $mo{l}^{-1}$ and $495$ kJ $mo{l}^{-1}$ respectively. The energy required to convert all the atoms present in $7$ mg of $Li$ vapours and $23$ mg of sodium vapours of their respective gaseous cations respectively are:

• A. $52$ J, $49.5$ J
• B. $520$ J, $495$ J
• C. $49.5$ J, $52$ J
• D. $495$ J, $520$ J

Answer 1

The correct option is B $520$ J, $495$ J

$IE$ of $Li=520Kj/mol$

$1$ mole of gaseous $Li$ requires $520KJ$ of energy to ionise each atom.

$1mol=6.02\times {10}^{23}$ atoms

So, $1$ atom=$\frac{1}{6.02times{10}^{23}}mol$

$520KJ=520000J=5.2\times {10}^{5}J$

$1mol⟶5.2\times {10}^{5}J$

$=\frac{7\times {10}^{-3}}{7}mol⟶5.2\times {10}^5\times {10}^{-3}J$

$=5.2\times {10}^{2}J$

$=520J$

$IE$ of $Na=495KJ/mol$

$1mol⟶495KJ$ energy

$=\frac{23\times {10}^{-3}}{23}mol⟶495\times {10}^{-3}KJ$

$=495J$

So, energy required to convert $Li$ and $Na$ to their respective gaseous cations are $520J$ and $495J$