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, \(5.00 \mathrm{~L}\) of \(\mathrm{N}_{2}\) gas is passed through \(7.2146 \mathrm{~g}\) of liquid benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\), at \(26.0{ }^{\circ} \mathrm{C}\). The liquid remaining after the experiment weighs \(5.1493 \mathrm{~g}\). 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?

First, we need to find the mass of benzene that evaporated during the experiment. This can be found by taking the difference between the initial mass and the final mass. $$\text{Mass of evaporated benzene} = \text{Initial mass} - \text{Final mass}$$ $$\text{Mass of evaporated benzene} = 7.2146\mathrm{~g} - 5.1493\mathrm{~g} = 2.0653\mathrm{~g}$$

To calculate the moles of evaporated benzene, we will use the molar mass of benzene which is approximately \(78.11\mathrm{~g/mol}\). $$\text{Moles of evaporated benzene} = \frac{\text{Mass of evaporated benzene}}{\text{Molar mass of benzene}}$$ $$\text{Moles of evaporated benzene} = \frac{2.0653\mathrm{~g}}{78.11\mathrm{~g/mol}} = 0.02645\mathrm{~mol}$$

Before we can use the Ideal Gas Law, we need to convert the given temperature from Celsius to Kelvin. $$T(K) = T(^{\circ}\mathrm{C}) + 273.15$$ $$T(K) = 26.0{ }^{\circ} \mathrm{C} + 273.15 = 299.15\mathrm{~K}$$

Now that we have the moles of evaporated benzene and the temperature in Kelvin, we can use the Ideal Gas Law to determine the vapor pressure (partial pressure) of benzene in the gaseous mixture. The Ideal Gas Law is as follows: $$PV = nRT$$ We'll rearrange the equation to find the partial pressure of benzene, where \(P\) is the partial pressure, \(n\) is the moles of evaporated benzene, \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, and \(V\) is the volume. $$P = \frac{nRT}{V}$$ For our calculation, we will use the gas constant \(R = 62.363\mathrm{~L\cdot torr/mol\cdot K}\). The given volume of the gas mixture is \(5.00\mathrm{~L}\). $$P_\mathrm{benzene} = \frac{(0.02645\mathrm{~mol})(62.363\mathrm{~L\cdot torr/mol\cdot K})(299.15\mathrm{~K})}{5.00\mathrm{~L}} = 98.91\mathrm{~torr}$$ Therefore, the vapor pressure of benzene in this experiment is approximately \(98.91\mathrm{~torr}\).