Question

First ionisation potential energy of Li(m) is 520 kJ/mol$^{-1}$, hence calculate the energy required to convert Li-atoms present in 70 mg Li(m) into Li$^+$ ion.

• A. 52 kJ
• B. 5.2 kJ
• C. 52 \(\times\) 10\(^3\) kJ
• D. 3640 kJ

Hint:

To calculate energy for a given mass, convert the mass to moles and then multiply by the ionisation energy per mole.

Answer 1

The Correct Option is A

The number of moles of Li atoms in 70 mg is: \[ \text{Moles of Li} = \frac{\text{Mass of Li}}{\text{Molar mass of Li}} = \frac{70}{6.94} = 10.08 \, \text{mmol}. \] The ionisation energy for 1 mole of Li atoms is 520 kJ/mol. So, the energy required for 10.08 mmol is: \[ \text{Energy} = 520 \times 10.08 \times 10^{-3} = 5.2 \, \text{kJ}. \]