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 N2 gas is passed through 7.2146 g of liquid benzene (C6H6) at 26.0 °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?

Verified step by step guidance

1. Calculate the mass of benzene that has evaporated by subtracting the final mass of the liquid from the initial mass: \( \text{mass evaporated} = 7.2146 \text{ g} - 5.1493 \text{ g} \).
2. Convert the mass of evaporated benzene to moles using its molar mass: \( \text{moles of C}_6\text{H}_6 = \frac{\text{mass evaporated}}{\text{molar mass of C}_6\text{H}_6} \).
3. Use the ideal gas law to find the partial pressure of benzene vapor. First, calculate the total moles of gas (N2 + benzene vapor) using the ideal gas law: \( PV = nRT \).
4. Determine the moles of N2 gas using the ideal gas law, assuming the initial conditions (before benzene evaporation) and subtract the moles of benzene vapor to find the moles of N2.
5. Calculate the vapor pressure of benzene by using the ratio of moles of benzene vapor to total moles of gas and multiplying by the total pressure: \( P_{\text{benzene}} = \left( \frac{\text{moles of C}_6\text{H}_6}{\text{total moles}} \right) \times P_{\text{total}} \).