How much liquid will be present?

Question

The vapor pressure of ethanol, CH₃CH₂OH, at 30.0 C is 10.46 kPa . If 2.22g of ethanol is enclosed in a 2.00 L container, how much liquid will be present?

Answer 1

moles present

n=(2.22g/46.08g)=0.04818mol

Moles that are gaseous:

PV=nRT

P=10.46kPa=0.10323217atm

V=2.00L

n=?

R=0.08206 L-atm/mol-K

T=30.0C=303.15K

n=PV/RT

n=(0.10323217*2.00)/(0.08206*303.15)

n=0.00829958mol

mass that is gaseous:

0.00829958mol(42.06g/1mol)=0.3824g

Mass liquid - mass gaseous = mass of liquid still present

2.22-0.3824=1.837555

=1.84g liquid