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...

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