Help With Basic Chem

Question

The vapor pressure of ethanol, CH3 CH2 OH, at 30.0 C is 10.46 kPa. If 2.27 g of ethanol is enclosed in a 2.00 L container, how much liquid will be present?

Answer 1

moles of ethanol present = 2.27 g x 1 mol/46.07 g = 0.04927 moles

PV = nRT

V = nRT/P

V = volume in literes = ?

n = moles = 0.04927

R = gas constant = 8.314 L-kPa/mol-L

T = temperature in Kelvin = 30.0 + 273.15 = 303.15K

P = pressure = 10.46 kPa

V = (0.04927)(8.314)(303.15) / 10.46

V = 11.87 L = 11.9 L (to 3 sig. figs.)