Name ID Section Last First Advance Study Assignment - Chemistry 180 Molar Mass o
ID: 693294 • Letter: N
Question
Name ID Section Last First Advance Study Assignment - Chemistry 180 Molar Mass of a Volatile Liquid Report all results to the correct number of significant figures An empty flask weighs 123.938 g. After vaporization of a sample of volatile liquid as described in the rinsed and filled completely with water at 21.8 "C. The mass of the flask is now 376.136g experimental write-up, at a measured temperature of 99.7 "C, the flask is sealed, cooled to room temperature, and weighed. The mass is now 124.205 g. The measured Pam is 757.3 torr. T 1. Calculate the volume of the flask from the mass of water contained in the flask and the density 2. Calculate the molar mass of the volatile liquid 169 Exp. 16-Molar MassExplanation / Answer
1) Assume that the density of water is 1 g/mL.
Mass of water contained in the flask = (mass of flask + water) = (mass of flask) = (376.136 – 123.938) g = 252.198 g.
Volume of water contained in the flask = volume of the flask = (mass of water in the flask)/(density of water) = (252.198 g)/(1 g/mL) = 252.198 mL = (252.198 mL)*(1 L/1000 mL) = 0.252198 L (a liquid fills up the entire volume of the container where it is placed) (ans).
2) Calculate the number of moles of the liquid vaporized using the ideal gas law.
Patm = 757.3 torr = (757.3 torr)*(1 atm/760 torr) = 0.99645 atm (1 atm = 760 torr).
Temperature of the gas = temperature of the water-bath = 99.7°C = (99.7 + 273) K = 372.7 K.
Use the ideal gas law:
P*V = n*R*T where V = 0.252198 L; plug in values and get
(0.99645 atm)*(0.252198 L) = n*(0.082 L-atm/mol.K)*(372.7 K)
====> n = 0.0082229 mole.
The number of moles of sample vaporized = 0.0082229 mole.
The mass of the sample vaporized = (124.205 – 123.938) g = 0.267 g.
Molar mass of the volatile liquid = (mass of sample vaporized)/(number of moles of sample vaporized) = (0.267 g)/(0.0082229 mol) = 32.4710 g/mol 32.471 g/mol (ans).
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