Question

A 255mL round-bottom flask is weighed and found to have a mass of 114.85g. A few mL of an easily vaporized liquid are added to the flask and the flask is immersed in a boiling water bath. All of the liquid vaporizes at the boiling temperature of water, filling the flask with vapor. When all of the liquid has vaporized, the flask is removed from the bath, cooled, dried and reweighed. The new mass of the flask and the condensed vapor is 115.23g. Which of the following compounds could the liquid be? (Assume the ambient pressure is 1.0atm)A. C2H5OHB. C4H10C. C2H6D. C4H9OHE. C3H7OH

Answer 1

Answer:

A) C2H5OH

Explanation:

The mass of the vapor (mv) will be the final mass less the mass of the flask:

mv = 115.23 - 114.85 = 0.38 g

Assuming the vapor as an ideal gas, we must use the ideal gas equation:

PV = nRT

Where P is the pressure, V the volume, n the number of moles, R the constant of the gases and T the temperature in Kelvin.

The liquid was boiled at 100ºC = 373 K at 1 atm.

R = 0.082 atm.L/(mol.K), and V = 255 mL = 0.255 L

1x0.255 = nx0.082x373

30.586xn = 0.255

n = 8.34x$10^{-3}$

Knowing that n = mass/ molar mass

Molar mass = 0.38/8.34x$10^{-3}$

Molar mass = 45.56 g/mol

By the periodic table, the molar mass of the elements given are: C = 12 g/mol; H = 1 g/mol; O = 16/mol. So the molar mass of the compounds are:

A) C2H5OH = 2x12 + 5x1 + 16 + 1 = 46 g/mol

B) C4H10 = 4x12 + 10x1 = 58 g/mol

C) C3H6 = 2x12 + 6x1 = 30 g/mol

D) C4H9OH = 4x12 + 9x1 + 16 + 1 = 74 g/mol

E) C3H7OH = 3x12 + 7x1 + 16 + 1 = 60 g/mol

Letter A is the closest.