Question

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?

Short answer

The vapor pressure of benzene is approximately $98.49\mathrm{torr}$.

Step-by-step solution

Calculate the mass of evaporated benzene

Subtract the final mass of liquid benzene from the initial mass to find the mass of benzene that has evaporated: Mass of evaporated benzene = Initial mass - Final mass Mass of evaporated benzene = $7.2146\mathrm{g}$ - $5.1493\mathrm{g}$ = $2.0653\mathrm{g}$

Determine the amount of benzene in moles

To calculate the amount of benzene in moles, use the molar mass of benzene, which is approximately $78.11\mathrm{g}/\mathrm{mol}$. Moles of benzene = Mass of evaporated benzene / Molar mass of benzene Moles of benzene = $2.0653\mathrm{g}$ / $78.11\mathrm{g}/\mathrm{mol}$ = $0.026451\mathrm{mol}$

Apply the ideal gas law

Use the ideal gas law formula, PV = nRT, to find the pressure of the benzene vapor. The values for R (the ideal gas constant) and T (temperature in Kelvin) are given. Convert the temperature from Celsius to Kelvin first: Temperature in Kelvin = $26.0{}^{\circ }\mathrm{C}+273.15=299.15\mathrm{K}$ Now, rearrange the ideal gas law equation to solve for pressure, P: P = nRT / V

Calculate the vapor pressure

Use the values of moles of benzene (n), temperature (T), volume of nitrogen gas (V), and the ideal gas constant (R) to find the vapor pressure of benzene: P = $(0.026451\mathrm{mol})(0.0821\mathrm{L}\cdot \mathrm{atm}/\mathrm{mol}\cdot \mathrm{K})(299.15\mathrm{K})/(5.00\mathrm{L})=0.1296\mathrm{atm}$

Convert the pressure to torr

In order to convert the vapor pressure of benzene from atm to torr, use the conversion factor $1\mathrm{atm}=760\mathrm{torr}$: Vapor pressure of benzene = $(0.1296\mathrm{atm})(760\mathrm{torr}/\mathrm{atm})=98.49\mathrm{torr}$ The vapor pressure of benzene is approximately $98.49\mathrm{torr}$.