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Question

The molar enthalpies of fusion and sublimation of molecular iodine are 15.2715.27 and 62.30 kJ mol162.30 \mathrm{~kJ} \mathrm{~mol}^{-1}, respectively. Estimate the molar heat of vaporization of liquid iodine.

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Given:

The molar enthalpy of fusion of iodine=15.27kJmol\mathrm{The \ molar \ enthalpy \ of \ fusion \ of \ iodine= 15.27 \dfrac{kJ}{mol}} The molar enthalpy of sublimation of iodine=62.30kJmol\mathrm{The \ molar \ enthalpy \ of \ sublimation \ of \ iodine= 62.30 \dfrac{kJ}{mol}}

The molar enthalpy of vaporization of iodine=?\mathrm{The \ molar \ enthalpy \ of \ vaporization \ of \ iodine=?}

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