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 (g) into $\mathrm{Li}^{2+}(\mathrm{g})$ $[\mathrm{Li}(\mathrm{g}) \rightarrow \mathrm{Li}^{2+}(\mathrm{g})]$ is _____ $\mathrm{kJ}\,\mathrm{mol}^{-1}$. (nearest integer) [Molar mass of $\mathrm{Li}=7\,\mathrm{g}\,\mathrm{mol}^{-1}$]

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Classification of Elements and Periodicity

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}

]

Answer: 4

Moles of lithium

= \frac{3.5 \times 1 0 ^{- 3} g}{7 g mo l ^{- 1}} = 5 \times 1 0^{- 4} mol

.
Converting

Li (g)

L i^{2 +} (g)

removes two electrons, so it requires the first and second ionisation enthalpies together:

Δ E = 520 + 7297 = 7817 kJ mo l^{- 1} .

Energy for the given amount

= 5 \times 1 0^{- 4} \times 7817 = 3.91 kJ

.
To the nearest integer this is

4

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Integer answer