Answer to Question #251627 in Organic Chemistry for Lilly

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 L of Nà ƒ ƒ ¢ ‚ ‚ gas is passed through 7.2146 g of liquid benzene (C6H6) at 26.0 °C. The liquid remaining after the experi ment weighs 5.1493 g. Assuming that the gas becomes satu rated 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

As given in the problem the initial mass of benznene = 7.3178 g

Mass of benzene left = 5.7378 g

So mass of benzene with which gas get saturated = 7.3178 - 5.7378 = 1.58g

so moles of benzene with which gas get saturated = mass / mol wt = 1.58 / 78.112 = 0.02022 moles

Temperature = 26.9 C = 26.9 + 273.15 = 300.05 K

Volume = 5.80 L

According to ideal gas equation

PV = nRT

P = pressure ; n = moles ; R = gas constant ; T = temperature ' V = volume

P X 5.80 = ( 0.02022 moles) (62.36 Torr-Litres / mol-K) (300.05 Kelvin)

P = 378.338 / 5.80 Torr

Vapour pressure of benzene = 65.230 Torr