Question

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`.

Answer 1

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...