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If lattice enthalpy and hydration enthalpy of $\mathrm{KCl}$ are $699 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $-681.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. What is the enthalpy of solution of $\mathrm{KCl}$ ?
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The correct answer is:
$17.20 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\begin{aligned} \Delta_{\text {soln }} \mathrm{H} & =\Delta_{\mathrm{L}} \mathrm{H}+\Delta_{\text {hyd }} \mathrm{H} \\ & =+699 \mathrm{~kJ} \mathrm{~mol}^{-1}+\left(-681.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\right) \\ & =17.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$
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