Question
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I need help with D and E part 1 and all of part 2 please. Thank you.
Experiment 9 Advance Study Assignment: Molar Mass of a Volatile Liquid 1. A student weighs an empty flask and stopper and finds the mass to be 54.868 g. She then adds about 5 ml of an unknown liquid and heats the flask in a boiling water bath at 100°C. Ater all the liquid is vapor- ized, she removes the flask from the bath, stoppers it, and lets it cool. After it is cool, she momentarily removes the stopper, then replaces it and weighs the flask and condensed vapor, obtaining a mass of 55.802 g. The volume of the flask is known to be 242.5 ml.. The barometric pressure in the laboratory that day is 747 mm Hg. What was the pressure of the vapor in the flask in atm? a. atm b. What was the temperature of the vapor in K? the volume of the flask in liters? L. c. What was the mass of vapor that was present in the flask? d. How many moles of vapor are present? moles e. What is the mass of one mole of vapor (Eq. 2)? MM g/mole How would each of the following procedural errors affect the results to be expected in this experiment? Give your reasoning in each case. 2. a All of the liquid was not vaporized when the flask was removed from the water bath. b. The student read the barometric pressure as 727 mm Hg rather than its true value of 747 m Hg c. The flask was left open to the atmosphere while it was being cooied, and the stopper was inserted just before the final weighing d. The fast was removed from the buch befone the wagor water All the liquid had vaporized had reached the temperature of the boiling
Explanation / Answer
1) (d)
according to ideal gas equation, PV = nRT
P =pressure = 747 mm Hg = 0.983 atm
V = volume = 242.5 ml = 242.5 / 1000 = 0.2425 L
T = temperature = 100oC = 100 + 273 = 373 K
n =number of moles
R = gas constant = 0.082 L atm / mol K
Thus, n = PV / RT = (0.983 atm)(0.2425 L) / (0.082 L atm / mol K) (373 K) = 0.0078 mole
(e) 0.0078 mole = 55.802 - 54.868 = 0.934 gm
we know that, 1 mole = molar mass of the molecule
again, moles = mass / molar mass
hence, molar mass = mass / moles = 0.934 gm / 0.0078 mole = 119.74 gm /mole
hence, mass of 1 mole of the vapour = 119.74 gm
