How much energy will be needed to convert 2.5 mol of Li atoms in the gaseous state to form Li (g)? First I.P of Li= 5.41 eV. 

I.P. of Li = 5.41 eV ${\mathrm{atom}}^{-}$

$\begin{array}{l}\times \frac{6.023\times {10}^{23}\times 1.602\times {10}^{-19}{\mathrm{Jmol}}^{-}}{1\mathrm{eV}{\text{ atom }}^{-}}\\ =522\times {10}^3\mathrm{J}{\mathrm{mol}}^{-}\end{array}$

$\therefore$ Energy needed to convert 2.5 mol Li (s) to Li (g) 

$\begin{array}{l}=2.5\mathrm{mol}\times 522\times {10}^3\mathrm{J}{\mathrm{mol}}^{-}\\ =1305\mathrm{kJ}\end{array}$