Pre-Lab Questions: (Typed answers submitted at the beginning of lab - 3 points)

Question

Pre-Lab Questions: (Typed answers submitted at the beginning of lab - 3 points) 1. Provide a summary of the procedure in your own words. Demonstrate that you understand what measurements you will be taking and what they will be used for. 2. Since you are measuring pressure in mmHg, which value of R should you use? 3. Suppose you have a stoppered flask containing CO2 gas at room temperature and pressure. If the volume of your stoppered flask is 300.0 mL, the pressure in the room is 770 mm Hg, and the temperature in your work area is 22.0°C, how many moles of CO2 are in the flask? 4. How many grams of CO2 are in the flask (described in part 3)? 5. Why do we measure the volume of the flask with water rather than using the volume listed on the side? 6. Why is the flask not firmly stoppered during sublimation?

Explanation / Answer

Parts (1) and (5) cannot be answered because you haven’t uploaded the experiment. I need to go over the experiment to answer these three questions. However, I can answer parts (2), (3), (4) and (6) for you.

(2) The unit of pressure is mmHg for the said experiment. We know that

760 mmHg = 1 atm.

Therefore, the unit of the gas constant, R should be in L-atm and the value of R chosen will be 0.082 L-atm/mol.K (ans)

(3) We shall employ the equation of state here. The following parameters are supplied.

Pressure, P = 770 mmHg = (770 mmHg)*(1 atm/760 mmHg) = 1.013 atm

Volume, V = 300 mL = (300 mL)*(1 L/1000 mL) = 0.3 L

Temperature = 22°C; we need to convert this temperature to the absolute value so that T = (273 + 22)K = 295 K

The value of the gas constant used is R = 0.082 L-atm/mol.K

The equation of state is

PV = nRT where n = moles of the gas; therefore,

(1.013 atm)*(0.3 L) = n.(0.082 L-atm/mol.K)*(295 K)

====> 0.3039 L-atm = n.(24.19 L-atm/mol)

====> n = (0.3039 L-atm)/(24.19 L-atm/mol) = 0.0126 mole 0.013 mole (ans)

(4) The molar mass of CO2 is 44 gm/mol. Therefore, weight of 0.013 mole of CO2 is

(0.013 mole)*(44 gm/1 mole) = 0.572 gm

Ans: The weight of CO2 is 0.572 gm

(6) Sublimation is the process where a solid directly changes into vapor or gas without passing through the liquid phase. If a solid is allowed to sublime in a glass vessel that’s firmly stoppered, then pressure will build up inside the closed vessel due to generation of the gas. The gas doesn’t have an outlet to escape and can lead to a tremendous pressure inside the vessel. The vessel can crack or explode under the pressure. Thus, it is always safe to leave the vessel unstoppered or loosely stoppered.

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