Question:
At $298 \mathrm{~K}$, the enthalpy of fusion of a solid $(\mathrm{X})$ is $2.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and the enthalpy of vaporisation of the liquid $(\mathrm{X})$ is $98.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The enthalpy of sublimation of the substance $(X)$ in $\mathrm{kJ} \mathrm{mol}^{-1}$ is__________ . (in nearest integer)
Solution:
$\Delta \mathrm{H}_{\text {sub }}=\Delta \mathrm{H}_{\text {fus. }}+\Delta \mathrm{H}_{\text {vap. }}$
$=2.8+98.2$
$=101 \mathrm{~kJ} / \mathrm{mol}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.