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$5.75 \mathrm{mg}$ of sodium vapour is converted to sodium ion. If the ionisation energy of sodium is $490 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and atomic weight is 23 units, the amount of energy needed for this conversion will be
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$0.1225 \mathrm{~kJ}$
$\mathrm{Na}(\mathrm{g}) \longrightarrow \mathrm{Na}^{+}(\mathrm{g})+\mathrm{e}^{-}$
$1 \mathrm{~mol} \mathrm{Na}(\mathrm{g})$
requires $490 \mathrm{~kJ} /$ mol for ionization.
$\therefore \quad \frac{5.75 \times 10^{-3}}{23}$ mol $\quad$ requires $490 \mathrm{~kJ} \times \frac{5.75 \times 10^{-3}}{23}=0.1225 \mathrm{~kJ}=122.5 \mathrm{~J}$
$1 \mathrm{~mol} \mathrm{Na}(\mathrm{g})$
requires $490 \mathrm{~kJ} /$ mol for ionization.
$\therefore \quad \frac{5.75 \times 10^{-3}}{23}$ mol $\quad$ requires $490 \mathrm{~kJ} \times \frac{5.75 \times 10^{-3}}{23}=0.1225 \mathrm{~kJ}=122.5 \mathrm{~J}$
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