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Give an answer for the follwoing exercise.

(a) Trouton's rule states that the ratio of the molar heat of vaporization of a liquid (ΔHvap )\left(\Delta H_{\text {vap }}\right) to its boiling point in kelvins is approximately 90 J/Kmol90 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}. Use the following data to show that this is the case and explain why Trouton's rule holds true.

Tbp(C)ΔHvap (kJ/mol) Benzene 80.131.0 Hexane 68.730.8 Mercury 35759.0 Toluene 110.635.2\begin{array}{lcc} & \boldsymbol{T}_{\mathbf{b p}}\left({ }^{\circ} \mathbf{C}\right) & \boldsymbol{\Delta} \boldsymbol{H}_{\text {vap }}(\mathbf{k} \mathbf{J} / \mathbf{m o l}) \\ \hline \text { Benzene } & 80.1 & 31.0 \\ \text { Hexane } & 68.7 & 30.8 \\ \text { Mercury } & 357 & 59.0 \\ \text { Toluene } & 110.6 & 35.2 \end{array}

(b) Use the values in the table to calculate the same ratio for ethanol and water. Explain why Trouton's rule does not apply to these two substances as well as it does to other liquids.

 Substance  Boiling point (C)ΔHvap (kJ/mol) Argon (Ar)1866.3 Benzene (C6H6)80.131.0 Ethanol (C2H5OH)78.339.3 Diethyl ether (C2H5OC2H5)34.626.0 Mercury (Hg2)35759.0 Methane (CH4)1649.2 Water (H2O)10040.79\begin{array}{ccc} \text { Substance } & \text { Boiling point }\left({ }^{\circ} \mathrm{C}\right) & \Delta H_{\text {vap }}(\mathrm{kJ} / \mathrm{mol}) \\ \text { Argon }(\mathrm{Ar}) & -186 & 6.3 \\ \text { Benzene }\left(\mathrm{C}_6 \mathrm{H}_6\right) & 80.1 & 31.0 \\ \text { Ethanol }\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right) & 78.3 & 39.3 \\ \text { Diethyl ether }\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5\right) & 34.6 & 26.0 \\ \text { Mercury }\left(\mathrm{Hg}^2\right) & 357 & 59.0 \\ \text { Methane }\left(\mathrm{CH}_4\right) & -164 & 9.2 \\ \text { Water }\left(\mathrm{H}_2 \mathrm{O}\right) & 100 & 40.79 \end{array}

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In this task, it should be explained why Trouton's rule holds true. The ratio for the ethanol and water should be calculated,

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