Question:
Lattice enthalpy and enthalpy of solution of $\mathrm{NaCl}$ are $788 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $4 \mathrm{~kJ} \mathrm{~mol}^{-1}$, respectively. The hydration enthalpy of $\mathrm{NaCl}$ is :
Correct Option: , 3
Solution:
$\Delta_{\text {sol. }} \mathrm{H}^{\circ}=\Delta_{\text {lattice }} \mathrm{H}^{\circ}+\Delta_{\text {Hyd. }} \mathrm{H}^{\circ}$
$4=788+\Delta_{\text {Hyd. }} \mathrm{H}^{\circ}$
$\Delta_{\text {Hyd. }} \mathrm{H}^{\circ}=-784 \mathrm{kJmol}^{-1}$
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