The vapor pressure of ethanol, CH3CH2OH , at 40.0 °C is 17.88 kPa . If 2.05 g of ethanol is enclosed in a 3.00 L container, how much liquid will be present?

Answer 1

Use the ideal gas law to find the moles of ethanol vapor:

PV = nRT

P = pressure = 17.88 kPa

V = volume = 3.00 L

R = gas constant = 8.314 LKPa/Kmol

T = temperature in Kelvin = 40C + 273 = 313K

n = moles = ?

Solving for moles (n)...

n = PV/RT = (17.88)(3) / (8.314)(313)

n = 0.0206 mols C₂H₅OH (ethanol)

mass of vapor = 0.0206 mols x 46.1 g C₂H₅OH / mol = 0.950 g C₂H₅OH vapor

Mass of remaining liquid = 2.05 g - 0.950 g = 1.10 g of liquid remain