First and second ionization enthalpies of lithium are $520 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $7297 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively.
Energy required to convert 3.5 mg lithium $(\mathrm{g})$ into $\mathrm{Li}^{2+}(\mathrm{g})\left[\mathrm{Li}(\mathrm{g}) \rightarrow \mathrm{Li}^{2+}(\mathrm{g})\right]$ is $\_\_\_\_$ kJ . (nearest integer) [Molar mass of $\mathrm{Li}=7 \mathrm{~g} \mathrm{~mol}^{-1}$ ]
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Solution
Number of moles of $\mathrm{Li}=\frac{3.5 \times 10^{-3}}{7}=5 \times 10^{-4}$
Total energy needed to convert $\mathrm{Li}(\mathrm{g}) \rightarrow \mathrm{Li}^{2+}(\mathrm{g})$ is
$$
5 \times 10^{-4}(520+7297)=3.9085 \mathrm{~kJ}=4 \mathrm{~kJ}
$$
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