The vapor pressure of a volatile liquid can be determined by slowly bubbling a k

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, 5.80 L of N2 gas is passed through 7.3178 g of liquid benzene, C6H6, at 26.9 C and atmospheric pressure. The liquid remaining after the experiment weighs 5.7378 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?

Explanation / Answer

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

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