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, 5.01 L of N2 gas is passed through 7.9286 g of liquid benzene, C6H6, at 27.3 ∘C and atmospheric pressure. The liquid remaining after the experiment weighs 5.9987 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?

Answer 1

Explanation:

It is given that the initial mass of benznene is 7.9286 g

Mass of benzene left = 5.9987 g

So, mass of benzene with which gas get saturated will be calculated as follows.

               = 7.9286 g  - 5.9987 g  = 1.9299 g

Therefore, moles of benzene with which gas get saturated = $\frac{mass}{ molar mass}$

                = $\frac{1.9299 g}{78.112 g/mol}$

                = 0.0247 moles

Temperature = $27.3^{o}C$ = 27.3 + 273.15 = 300.45 K

Volume = 5.01 L

So, according to ideal gas equation PV = nRT

Putting the given values into the ideal gas equation as follows.

                       PV = nRT

           $P \times 5.01 L$ = $0.0247 mol \times 62.36 torr-liter/mol K \times 300.45 K$

                   P = $\frac{462.781 torr-liter}{5.01 L}$

                      = 92.371 torr

Hence, we can conclude that vapor pressure of benzene is 92.371 torr.