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At 298 \mathrm{~K}, the enthalpy of fusion of a solid (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 (\mathrm{X})$ in $\mathrm{kJ} \mathrm{mol}^{-1} is__________. (in nearest integer)
 

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best_answer

Given,

\mathrm{X\left ( s \right )\rightarrow X\left ( l \right ),\Delta H_{fus}= 2.8\ KJ\, mol^{-1}}
\mathrm{X\left ( l \right )\rightarrow X\left ( g \right ),\Delta H_{vap}= 98.8KJ\, mol^{-1}}

Now,  enthalpy of sublimation is defined as the enthalpy change when 1 mole of a solid is converted to its gaseous state

\mathrm{X\left ( S \right )\rightarrow X\left ( g \right )}

\mathrm{\therefore \Delta H_ {sub}= \Delta H_{fus}+\Delta H_{vap}}
\therefore \mathrm{\Delta H_{sub}= 2.8+98.2 = 101 \ kJ\ mol^{-1}}

Hence, the answer is 101

Posted by

sudhir.kumar

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