# 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$$/$$c {m}^{3}$.

##### 1 Answer
May 27, 2018

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

Here...${P}_{\text{solution}} = 50 \cdot m m \cdot H g$...

#### Explanation:

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

${\chi}_{\text{water"="moles of water"/"moles of water + moles of glycerol}}$

${\chi}_{\text{water}} = \frac{\frac{338 \cdot m L \times 0.992 \cdot g \cdot m {L}^{-} 1}{18.01 \cdot g \cdot m o {l}^{-} 1}}{\frac{338 \cdot m L \times 0.992 \cdot g \cdot m {L}^{-} 1}{18.01 \cdot g \cdot m o {l}^{-} 1} + \frac{164 \cdot g}{92.09 \cdot g \cdot m o {l}^{-} 1}} = \frac{18.62 \cdot m o l}{18.62 \cdot m o l + 1.781 \cdot m o l} = 0.913$

${\chi}_{\text{glycerol}} = \frac{\frac{164 \cdot g}{92.09 \cdot g \cdot m o {l}^{-} 1}}{\frac{338 \cdot m L \times 0.992 \cdot g \cdot m {L}^{-} 1}{18.01 \cdot g \cdot m o {l}^{-} 1} + \frac{164 \cdot g}{92.09 \cdot g \cdot m o {l}^{-} 1}} = \frac{1.781 \cdot m o l}{18.62 \cdot m o l + 1.781 \cdot m o l} = 0.0873$

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. ${\chi}_{\text{water"+chi_"glycerol}} \equiv 1$

And so $\text{vapour pressure"=chi_"water} \times 54.74 \cdot m m \cdot H g$

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

As is typical for aqueous solutions...${\chi}_{\text{water}}$ DOMINATES....the molar mass of water is so low...