Q.
For one mole of $NaCl_{(s)}$ the lattice enthalpy is
Solution:
$\Delta /H° = S + D +J.E.+ E.A. + U $
$- 411.2 = 108.4 + 121 + 495.6 - 348.6 + U$
$U = - 787.6\, kJ/mol$
$= - 788 \,kJ/mol$