First and second ionization enthalpies of lithium are 520 kJ mol$$^{-1}$$ and 7297 kJ mol$$^{-1}$$ respectively. Energy required to convert 3.5 mg lithium (g) into Li$$^{2+}$$(g) [Li(g) $$\to$$ Li$$^{2+}$$(g)] is _______ kJ mol$$^{-1}$$. (nearest integer)
[Molar mass of Li = 7 g mol$$^{-1}$$]

Given:

$$I\ E_1\ of\ Li\ =\ 520\ kJ\ mol^{-1}$$

$$I\ E_2\ of\ Li\ =\ 7297\ kJ\ mol^{-1}$$

Reaction:

$$Li\left(g\right)\longrightarrow\ Li^{+2}\left(g\right)+2e^-$$

Total energy required for one mole:

$$=I\ E_1+I\ E_2$$

$$=520+7297$$

$$=7817\ kJ\ mol^{-1}$$

Given Mass of Li: $$3.5\ \times\ 10^{-3}g$$

Moles of Li :

$$=\frac{3.5\ \times\ 10^{-3}g}{7}$$

$$=5\times\ 10^{-4}mol$$

Therefore energy required is:

$$=7817\times\ 5\times\ 10^{-4}$$

$$=3.9085\ kJ$$

Nearest Integer is 4.