Glycerin, #C_3H_8O_3#, is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 164 g glycerin to 338 mL #H_2O# at 39.8°C?

The vapor pressure of pure water at 39.8°C is 54.74 torr and its density is .992 #g##/##cm^3#.

1 Answer
anor277
May 27, 2018

The vapour pressure of the solution is proportional to the mole fraction of the volatile component...

Here...#P_"solution"=50*mm*Hg#...

Explanation:

And so the vapour pressure will be somewhat reduced with respect to PURE water...we calculate the mole fractions of each component...

#chi_"water"="moles of water"/"moles of water + moles of glycerol"#

#chi_"water"=((338*mLxx0.992*g*mL^-1)/(18.01*g*mol^-1))/((338*mLxx0.992*g*mL^-1)/(18.01*g*mol^-1)+(164*g)/(92.09*g*mol^-1))=(18.62*mol)/(18.62*mol+1.781*mol)=0.913#

#chi_"glycerol"=((164*g)/(92.09*g*mol^-1))/((338*mLxx0.992*g*mL^-1)/(18.01*g*mol^-1)+(164*g)/(92.09*g*mol^-1))=(1.781*mol)/(18.62*mol+1.781*mol)=0.0873#

I did this the long way, but of course we know that the SUM of the mole fractions in a BINARY solution is UNITY...i.e. #chi_"water"+chi_"glycerol"-=1#

And so #"vapour pressure"=chi_"water"xx54.74*mm*Hg#

#=underbrace0.913_(chi_"water")xx54.74*mm*Hg=??*mm*Hg#

As is typical for aqueous solutions...#chi_"water"# DOMINATES....the molar mass of water is so low...

Related questions
See all questions in Colligative Properties
Impact of this question
16168 views around the world
You can reuse this answer
Creative Commons License