From the time the mass of the flask is first measured in Part A.1 until the time

Question

From the time the mass of the flask is first measured in Part A.1 until the time it is finally measured in Part B.3, it is handled a number of times with oily fingers. Does this lack of proper technique result in the molar mass of the vapor in the flask being reported as too high or too low or as unaffected? Explain. The aluminum foil is pierced several times with large pencil-size holes instead of pin-size. How will this oversight in the procedure affect the mass of vapor measured in Part B.3, too low, too high, or unaffected? Explain. Will the reported molar mass of the liquid be reported too low, too high, or unaffected? Explain. The flask is completely filled with vapor only when it is removed from the hot water bath in Part B.3. However, when the flask cools, some of the vapor condenses in the flask. As a result of this observation, will the reported molar mass of the liquid be too high, too low, or unaffected? Explain.

Explanation / Answer

Ans. Part A1. Presence of oil from fingers on the external walls of the flask slightly increases the mass of the flask. It finally yields “relatively higher mass of the vapor”.

Moles of vapor is calculated from vapor pressure (P) and volume of flask (V) using ideal gas equation.

Now,

Molar mass = Mass of vapor / moles of vapor           - equation 1

Since, oil from fingers increases ‘mass of vapor in the flask’ after calculation, the apparent/ observed molar mass is also increased.

Part A2. a. The purpose of aluminum foils with few pinholes is to provide a small surface area through which excess vapor escape the flask when the liquid form is completely vaporized in water bath.   

Presence of numerous large pencil-size holes provide a large surface area of diffusion through which larger amount of vapor escape to equilibrate itself to the atmospheric pressure. So, finally there would less vapor molecules retained in the flask. So, mass of vapor would be lower than the actual.

b. Lower mass of vapor due to fewer molecules of vapor also give “relatively low vapor pressure”. So, there is lowering of both the mass and pressure in the vapor.

Using ideal gas equation, number of moles would be “lesser than actual” because observed pressure is comparatively low.   

Again, using equation 1, lower mass of vapor and lower number of moles would leave the molar mass “unaffected”- it’s assumed that no atmospheric air enters into the flask

Part B2. Condensation of vapor into liquid reduces the pressure in the flask. Calculating the moles of vapor using ideal gas equation gives “relatively fewer moles” due to apparently low pressure.

However, the mass of flask remains the same- whether the molecule is present in liquid phase or vapor phase or a mixture of both- because there is no net gain or loss of molecules.

So, Mass of vapor remains the same whereas “number of moles is reduced”.

Using equation 1, it would give “relatively higher molar mass”.

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