Toluene (C6H5CH3) is a liquid used in TNT. Its normal boiling point is 111.0 degrees Celsius, and its molar heat of vaporization is 35.9 kJ/mol. What would be the vapor pressure of toluene at 111.0 degrees Celsius? Explain.
Added by Joshua C.
Step 1
9 kJ/mol = 35,900 J/mol), R is the ideal gas constant (8.314 J/(mol*K)), T1 is the initial temperature (384 K), and T2 is the new temperature (348 K). Show more…
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Toluene, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}$, is a liquid used in the manufacture of TNT. Its normal boiling point is $111.0^{\circ} \mathrm{C}$, and its molar heat of vaporization is $35.9 \mathrm{~kJ} / \mathrm{mol}$. What would be the vapor pressure (torr) of toluene at $75.00^{\circ} \mathrm{C}$ ?
The normal boiling point of benzene is $80.1^{\circ} \mathrm{C},$ and the heat of vaporization is $\Delta H_{\mathrm{vap}}=30.7 \mathrm{kJ} / \mathrm{mol}$ . What is the boiling point of benzene in $^{\circ} \mathrm{C}$ on top of Mt. Everest, where $P=260 \mathrm{mm} \mathrm{Hg}$ ?
A benzene-toluene solution with $x_{\text {benz }}=0.300$ has a normal boiling point of $98.6^{\circ} \mathrm{C} .$ The vapor pressure of pure toluene at $98.6^{\circ} \mathrm{C}$ is $533 \mathrm{mm} \mathrm{Hg}$. What must be the vapor pressure of pure benzene at $98.6^{\circ} \mathrm{C} ?$ (Assume ideal solution behavior.)
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