The lattice energy of solid NaCl is 180 K. Cal per mol. The dissolution of the solid in water in the form of ions is endothermic to the extent of 1K. Cal per mol. If the solvation energies of ${\mathrm{Na}}^{+} \mathrm{and} {\mathrm{Cl}}^{-}$ ions are in ratio 6:5, what is the enthalpy of hydration of sodium ion?

.The enthalpy change for dissolution is the sum of the lattice enthalpy and the enthalpy change for hydration

ΔH_sol​=ΔH_lattice​+ΔH_hyd​

Substitute values in the above expression,

1 = 180 + ΔH_hyd​
ΔH_hyd ​= −179 kcal

The enthalpy of hydration for sodium chloride is the sum of the enthalpies of hydration for sodium ions and chloride ions

ΔH_Na+_(h)​+ΔH_Cl^--(h)​ = − 179 kcal

The hydration energies of sodium ion and chloride ions are in the ratio 6:5.
∴ΔH_Na+(h)​=$\frac{6 x -179}{11}=-97.64 k cal$