1.a) 2.5 mL of acetone was placed in a 250 mL round bottom flask equipped with a

Question

1.a) 2.5 mL of acetone was placed in a 250 mL round bottom flask equipped with a one-hole stopper and a glass tubing that extends into the flask. The liquid was evaporated using a boiling water bath at 100 °C. The excess vapor escaped through the pinhole of the tube until no visible liquid remained in the flask. The atmospheric pressure was measured 745 torr with a barometer in the room.

The mass of the flask, one-hole stopper and the glass tubing, and vapor was 94.58 g.

The initial mass of the dry, empty flask, one-hole stopper and the glass tubing was 94.12 g.

What is the molecular weight of a volatile liquid?

Explanation / Answer

Ans. Given,

Temperature, T = 1000C = 373.15 K

Volume of flask = 250.0 mL = 0.250 L

Mass of vapor = 94.58 g – 94.12 g = 0.46 g

Pressure = 745 torr = 0.98 atm                                          ; [1 torr = 0.00131579 atm]

It’s assumed that the vapor behaves ideally under the given conditions.

Ideal gas equation: PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol-1K-1

            T = absolute temperature (in K) = (0C + 273.15) K

Putting the values in equation 1-

            0.98 atm x 0.250 L = n x (0.0821 atm L mol-1K-1) x 373.15 K

            Or, n = (0.245 atm L) / (30.635615 atm L mol-1) = 0.007997 mol

Therefore, number of moles of vapor in the flask = 0.007997 mol

Now,

Molar mass = Mass / Number of moles

                        = 0.46 g / 0.007997 mol

                        = 57.52 g/ mol

Therefore, molar mass of the volatile liquid = 57.52 g/ mol

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