[MyLittlePony]

A stream of air is bubbled slowly through liquid benzene in a flask at 20 degrees celsius against an ambient pressure of 100.56 kPa. After the passage of 4.80 L of air (measured at 20 degrees and 100.56 kPa before it contains benzene vapor), it is found that 1.705 g of benzene have been evaporated. Assuming that the air is saturated with benzene, calculate the equilibrium vapor pressure of benzene at 20 degrees.

Relevant equations:

RTln(P/Po) = Vm(P - Po)

The correct answer is 10 kPa, but I'm confused as to how this is. Am I using the right equation?

[freezard7734]

Hi. Actually, the problem requires no special equations. All you need is the ideal gas law.
The only relevant information is the volume of air, the R constant, the temperature, and the ambient pressure to calculate the total moles of air molecules bubbled through the benzene. With that, and the 1.705 g of benzene (C6H6: MM = 78 g/mol), you can calculate the vapor pressure of benzene afterwards:

n_air = P * V_air / RT = .198 moles
n_benzene = m_benzene / MM_benzene = 1.705 / 78 = .021 moles

P_benzene = P_air * n_benzene / n_air = 100.56 * .021 / .198 = 10 kPa