Question
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.)
Step 1
We know that the mole fraction of benzene is $0.300$. Since the sum of the mole fractions of all components in a solution is equal to 1, the mole fraction of toluene is $1 - 0.300 = 0.700$. Show more…
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Benzene and toluene form nearly ideal solutions. The boiling point of pure benzene is 80.1°C. Calculate the chemical potential of benzene relative to that of pure benzene when xbenzene = 0.30 at its boiling point. If the activity coefficient of benzene in this solution were actually 0.93 rather than 1.00, what would be its vapor pressure?
The normal boiling point of the benzene-toluene mixture, where the molar fraction of benzene is 0.30, is 98.6 °C. The vapor pressure of pure toluene at 98.6 °C is 533 mmHg. What is the vapor pressure of pure benzene at 98.6 °C? (Assume the solution is ideal)
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