At 298 K, the enthalpy of fusion of a solid (X) is 2.8 kJ mol^$-$1 and the enthalpy of vaporisation of the liquid (X) is 98.2 kJ mol^$-$1. The enthalpy of sublimation of the substance (X) in kJ mol^$-$1 is _____________. (in nearest integer)

<p>The enthalpy of sublimation is the total amount of energy required to convert a solid directly into a gas. This can be calculated by summing the enthalpy of fusion (solid to liquid) and the enthalpy of vaporization (liquid to gas). Mathematically, this relationship is represented as:</p> <p>$$ \Delta H_{\text{sublimation}} = \Delta H_{\text{fusion}} + \Delta H_{\text{vaporization}} $$</p> <p>Given:</p> <p><p>Enthalpy of fusion, $\Delta H_{\text{fusion}} = 2.8 \, \text{kJ mol}^{-1}$</p></p> <p><p>Enthalpy of vaporization, $\Delta H_{\text{vaporization}} = 98.2 \, \text{kJ mol}^{-1}$</p></p> <p>Substitute these values into the equation:</p> <p>$$ \Delta H_{\text{sublimation}} = 2.8 \, \text{kJ mol}^{-1} + 98.2 \, \text{kJ mol}^{-1} = 101.0 \, \text{kJ mol}^{-1} $$</p> <p>Therefore, the enthalpy of sublimation of the substance (X) is approximately 101 kJ mol^$-1$.</p>