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

Solution :
No. of moles of $$Li = \frac{7}{{1000 \times 7}} = {10^{ - 3}}$$
No. of moles of $$Na = \frac{{23}}{{1000 \times 23}} = {10^{ - 3}}$$
∴ The amount of energies required for $${10^{ - 3}}$$  mole each of $$Li$$  and $$Na$$  are $$520\,kJ \times {10^{ - 3}}$$   and $$495\,kJ \times {10^{ - 3}}$$   or $$520\,J$$  and $$495\,J$$  respectively.