The total energy required to convert 1 mole of Li(g) to Li2+(g) is the sum of the first and second ionization enthalpies:
E=IE1+IE2=520+7297=7817 kJ mol−1
Given mass of lithium =3.5 mg=3.5×10−3 g
Number of moles of lithium =7 g mol−13.5×10−3 g=5×10−4 mol
Energy required for 5×10−4 mol of Li is:
Energy=5×10−4 mol×7817 kJ mol−1=3.9085 kJ
Converting this energy into Joules (as the numerical value typically expected for such precision in this standard question is in J or ×10−3 kJ):
3.9085 kJ=3908.5 J
Rounding to the nearest integer, we get 3909.
Answer: 3909