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, 4.17 L of N2 gas is passed through 7.0150 g of liquid benzene, C6H6, at 27.0 C and atmospheric pressure. The liquid remaining after the experiment weighs 5.5149 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

Solution :-

Lets calculate the difference in the mass of the benzene

7.0150 g – 5.5149 g = 1.5001 g

Now lets calculate the moles of the benzene

Moles of benzene = 1.5001 g / 78.11 g peer mol = 0.019205 mol benzene

We know the gas is saturated with benzene.

27 C +273 = 300 K

V= 4.17 L

PV= nRT

P= nRT/V

= 0.019205 mol * 0.08206 L atm per mol K * 300 K / 4.17 L

= 0.1134 atm

So the vapor pressure of the benzene is 0.1134 atm

We can convert it into torr as

0.1134 atm * 760 torr / 1 atm = 86.2 torr

So the vapor pressure of the benzene is 86.2 torr

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