The vapor pressure of a volatile liquid can be determined by slowly bubbling a known volume of gas through it at a known temperature and pressure. In an experiment, 5.00 L of \(\mathrm{N}_{2}\) gas is passed through 7.2146 \(\mathrm{g}\) of liquid benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) at \(26.0^{\circ} \mathrm{C} .\) The liquid remaining after the experiment weighs 5.1493 g. Assuming that the gas becomes saturated with benzene vapor and that the total gas volume and temperature remain constant, what is the vapor pressure of the benzene in torr?

Firstly, we need to find how much benzene was vaporized in the experiment. We can do this by subtracting the final mass of benzene from the initial mass. Mass of vaporized benzene = Initial mass - Final mass Mass of vaporized benzene = 7.2146 g - 5.1493 g = 2.0653 g

Now convert the mass of vaporized benzene to moles using the molar mass of benzene, which is 78.11 g/mol for C6H6. Moles of vaporized benzene = (Mass of vaporized benzene)/(Molar mass of benzene) Moles of vaporized benzene = (2.0653 g)/(78.11 g/mol) = 0.02645 mol

Use the ideal gas law equation (PV = nRT) to find the initial pressure of the nitrogen gas. We are given the volume of gas (5.00 L) and temperature (\(26^{\circ} \mathrm{C}\); converted to Kelvin as \(26 + 273.15 = 299.15K\)). We know that R, the ideal gas constant, is 0.0821 L atm/mol K. First, we need to find the moles of nitrogen gas. To do this, we can use nitrogen's molar mass (28.02 g/mol) and the fact that nitrogen makes up 79% of the atmosphere by volume (assuming other gases do not contribute to the pressure). Then, we can find the atmospheric pressure P in atm by using the fact that the atmospheric pressure is 760 torr. Because the nitrogen gas is 79% by volume, we can calculate the initial pressure of nitrogen gas as follows. Atmospheric pressure = 760 torr = 1 atm (approximately) Initial pressure of nitrogen = 0.79 * Atmospheric pressure Initial pressure of nitrogen = 0.79 * 1 atm = 0.79 atm Now we can find the moles of nitrogen gas (n). PV = nRT n = PV/(RT) n = (0.79 atm * 5.00 L)/ (0.0821 L atm/mol K * 299.15 K) = 0.1617 mol

Next, we need to find the total moles of gas present after the nitrogen gas is saturated with benzene vapor. We can do this by adding the moles of nitrogen gas and the moles of vaporized benzene. Total moles of gas = Moles of nitrogen gas + Moles of vaporized benzene Total moles of gas = 0.1617 mol + 0.02645 mol = 0.18815 mol

Now we need to find the new pressure experienced by the gas mixture using the ideal gas law equation. PV = nRT P = nRT/V P = (0.18815 mol * 0.0821 L atm/mol K * 299.15 K)/(5.00 L) = 0.9887 atm

Finally, let's find the vapor pressure of benzene, which can be determined by finding the partial pressure of benzene in the gas mixture. We can do this by subtracting the initial pressure of nitrogen gas from the new pressure. Partial pressure of benzene vapor = New pressure - Initial pressure of nitrogen Partial pressure of benzene vapor = 0.9887 atm - 0.79 atm = 0.1987 atm Now we need to convert this pressure to torr. Vapor pressure of benzene = 0.1987 atm * (760 torr/1 atm) = 150.8 torr The vapor pressure of benzene is approximately 150.8 torr.