The vapor pressure of ethanol, CH3CH2OH , at 35.0 °C is 13.67 kPa . If 2.01 g of ethanol is enclosed in a 2.50 L container, how much liquid will be present?

Answer 1

First, we'll find the mass of ethanol in the vapor phase, and then we'll subtract that from the original mass of ethanol present (2.01g) to solve for mass of ethanol in the liquid phase.

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

PV = nRT

P = pressure = 13.67 kPa

V = volume = 2.50 L

n = moles of vapor = ?

R = gas constant = 8.314 L-kPa/Kmol

T = temperature in K = 35.0º + 273 = 308K

Solving for n we have ...

n = PV / RT = (13.67)(2.50) / (8.314)(308)

n = 0.01335 moles of vapor

Convert moles of vapor to grams of vapor using the molar mass of ethanol (46.07 g/mol):

0.01335 mols x 46.07 g / mol = 0.6148 g of vapor

Mass of remaining liquid = 2.01 g - 0.615 g = 1.40 g of liquid present