You are given a quantity of a volatile (easily vaporized) liquid substance and a

Question

You are given a quantity of a volatile (easily vaporized) liquid substance and asked to determine its molecular weight. To do that, you place a sample of the liquid in a 200.0 mL flask and submerge the flask in boiling water to vaporize all of the liquid. The vapor fills the flask and the excess escapes. Now you cool the flask so that all the vapor in it condenses, and you weigh the flask and liquid. After you subtract the mass of the flask, you find that the mass of the sustance in the flask is 0.970 g. The temperature of the boiling water was 99 degrees celsius; the barometric pressure was 733 mm Hg. What is the molecular weight of the substance? (Ideal gas equation, MW and density form)

Explanation / Answer

Mass of the substance = 0.970 g.

Since the vaporized substance occupies the entire space inside the flask (volume of the flask), hence the volume of the vaporized substance = 200 mL = 0.200 L.

Therefore, density of the vapor = mass of vapor/volume of vapor = 0.970 g/0.200 L = 4.85 g/L.

Pressure of the vapor = 733 mmHg = (733 mmHg)*(1 atm/760 mmHg) = 0.96447 atm.

Temperature of the vapor = temperature of the water bath = 99C = (273 + 99) K = 372 K.

Use the ideal gas law:

PV = nRT where n = m/M is the moles of the vapor; m is the mass of the vaporized substance and M is the molecular weight of the substance.

Therefore,

P = (m/MV)RT = (/M)RT where = m/V is the density of the substance.

Therefore,

M = (/P)RT

Plug in values and write

M = (4.85 g/L)*(0.082 L-atm/mol.K).(372 K)/(0.96477 atm) = 153.3468 g/mol 153.35 g/mol.

The molecular weight of the substance is 153.35 g/mol (ans).

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