# What is the vapor pressure in mmHg of a solution of 16.0 g of glucose (C6H12O6) in 80.0 g of methanol (CH3OH) at 27 degrees C? The vapor pressure of pure methanol at 27 degrees C is 140 mmHg.

Dec 14, 2015

${P}_{\text{sol" = "135 mmHg}}$

#### Explanation:

In order to be able to determine the vapor pressure of this solution, you need to know two things

• the vapor pressure of pure methanol at ${27}^{\circ} \text{C}$ - given to you as $\text{140 mmHg}$
• the mole fraction of methanol in the solution

As you know, the vapor pressure of a solution that contains a non-volatile solute will depend exclusively on the mole fraction of the solvent, which in your case is methanol, and on the pure solvent's vapor pressure at that temperature

$\textcolor{b l u e}{{P}_{\text{sol" = chi_"solvent" xx P_"solvent}}^{\circ}}$

The mole fraction of the solvent will be equal to the number of moles of solvent divided by the total number of moles present in solution.

Use the molar masses of the two compounds to determine how many moles of each you have

16.0 color(red)(cancel(color(black)("g"))) * ("1 mole C"_6"H"_12"O"_6)/(180.16color(red)(cancel(color(black)("g")))) = "0.08881 moles C"_6"H"_12"O"_6

80.0color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"OH")/(32.04color(red)(cancel(color(black)("g")))) = "2.4969 moles CH"_3"OH"

The total number of moles present in solution will be

${n}_{\text{total" = n_"glucose" + n_"methanol}}$

${n}_{\text{total" = 0.08881 + 2.4969 = "2.5857 moles}}$

The mole fraction of methanol will be

chi_"methanol" = (2.4969 color(red)(cancel(color(black)("moles"))))/(2.5857color(red)(cancel(color(black)("moles")))) = 0.9657

The vapor pressure of the solution will thus be

${P}_{\text{sol" = 0.9657 * "140 mmHg" = "135.2 mmHg}}$

I'll leave the answer rounded to three sig figs, despite the fact that you only have two sig figs for the vapor pressure of the pure solvent

P_"sol" = color(green)("135 mmHg")