Question

Acetone (C₃H₆O) is a common solvent used in organic chemistry. Its normal boiling point is 56.5^oC. Acetone has an enthalpy of vaporization of 31.3 kJ/mol. Estimate what mass of acetone would be present in 1.0 L of air at 25^oC that is saturated with acetone. (Hint: you will need to determine the vapor pressure of acetone at 37^oC).

Answer 1

Vapor pressure of acetone at 25 deg.c need to be determied based on the information on latent heat of vaporization.

Classisus clayperon equation can be used for this purpose

ln (P2/P1)= (delH/R)*(1/T1-1/T2)

where P2= Vapor pressure at T2= 56.5 deg.c =56.5+273.15=329.65K

P2= 760 mm Hg ( Normal boiling point refers to a pressure of 760 mm Hg= 1 atm)

P1= vapor pressure at T1= 25 deg.c =25+273.15= 298.15K

delH= latent heat of vaporization= 31.3Kj/mol= 31.3*1000 joules/ mole and R= 8.314 j/mol

ln(760/P2)= (31.3*1000/8.314)*(1/298.15-1/329.65)

760/P2= 3.34

P2= 760/3.34=227.5 mm Hg

Since the air is saturated with this acetone at the given temperature, this vapor pressure is also equal to the partial pressure

partial pressure = 227.5 mm Hg=227.5/760=0.060 atm (partial pressure is the pressure exerted by the gas it it alonee occupies the entire volume)

T=25 deg.c =25+273.15= 298.15 Volume, V =1 L from PV= nRT, R =0.08206 L.atm/mole.K

n= number of moles = PV/RT= 0.060*1/(0.08206*298.15)=0.002452 moles

molecular weight of acetone (C3H6O)= 3*12+6+16= 58

Mass of acetone = moles* Molecular weight of acetone = 0.002452*58=0.1422 gms