The vapor pressure of ethanol, CH3CH2OH , at 35.0 °C is 13.67 kPa . If 2.14 g of ethanol is enclosed in a 2.50 L container, how much liquid will be present?

Answer 1

moles CH3CH2OH = 2.14 g x 1 mol / 46.08 g = 0.04644 moles

moles CH3CH2OH that are in the gas phase :

PV = nRT

n = PV/RT = (13.67 kPa)(2.50 L) / (8.314 LkPa/Kmol)(308K)

n = 0.01335 moles in the gas phase

moles CH3CH2OH in the liquid phase = total moles - moles in gas phase

0.04644 moles - 0.01335 moles = 0.03309 moles in the liquid phase

mass of liquid = 0.03309 moles x 46.08 g / mol = 1.52 g ethanol in the liquid phase