At 88 , ^circ C,the vapor pressure of benzene | vedclass

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Question

(D) When the mixture of benzene and toluene is boiled at $88 \, ^\circ C$,the total vapor pressure of the solution becomes equal to the external press

Vedclass · Accepted Answer

$0.740$

Vedclass · Answer

(D) When the mixture of benzene and toluene is boiled at $88 \, ^\circ C$,the total vapor pressure of the solution becomes equal to the external pressure,which is $1 \, 	ext{atm} = 760 \, 	ext{torr}$.According to Raoult's Law for an ideal solution: $P_S = P_A^0 X_A + P_B^0 X_B$Here,$P_S = 760 \, 	ext{torr}$,$P_A^0 = 900 \, 	ext{torr}$ (benzene),$P_B^0 = 360 \, 	ext{torr}$ (toluene),and $X_B = 1 - X_A$.Substituting the values: $760 = 900 X_A + 360(1 - X_A)$$760 = 900 X_A + 360 - 360 X_A$$760 - 360 = 540 X_A$$400 = 540 X_A$$X_A = \frac{400}{540} \approx 0.7407 \approx 0.740$.

At $88 \, ^\circ C$,the vapor pressure of benzene is $900 \, \text{torr}$ and the vapor pressure of toluene is $360 \, \text{torr}$. What is the mole fraction of benzene in the mixture with toluene at $1 \, \text{atm}$ and $88 \, ^\circ C$? Assume benzene and toluene form an ideal solution.

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