A mixture of benzene and toluene has a total vapor pressure at 25 °C of 45.06 mmHg.What is the partial pressure of benzene in this solution? The vapor pressure of pure benzene and pure toluene at 25 °C are 95.03 mmHg and 28.40 mmHg respectively.
A)0.7500 mmHg
B)23.76 mmHg
C)0.3217 mmHg
D)21.30 mmHg
E)16.66 mmHg

Question

Quizplus · Accepted Answer

The partial pressure of benzene can be calculated using Raoult's law, which states that the partial pressure of a component in a mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture. Given the total vapor pressure and the vapor pressures of the pure components, we can set up an equation to solve for the mole fraction of benzene (x_benzene) and then calculate its partial pressure. However, without the mole fractions, we can't directly calculate it here. The provided information is not sufficient to directly calculate the partial pressure without assuming the mole fractions or using additional information. The correct answer, however, is based on understanding that the total vapor pressure is a result of the contributions from both components and knowing that the vapor pressure of benzene is significantly higher than that of toluene, indicating that benzene would have a substantial contribution to the total vapor pressure. The choice provided here as correct (B) is likely based on an error in the question's setup or a misunderstanding, as the exact calculation requires more information or a different approach.

A Mixture of Benzene and Toluene Has a Total Vapor