How would you determine the vapor pressure of a solution at 25C that contains 76.6 g of glucose (C6H12O6) in 250.0 mL of water? The vapor pressure of pure water at 25C is 23.8 torr.

Nov 11, 2015

$\text{23.1 torr}$

Explanation:

The first thing to do here is determine the mass of the sample of water by using water's density at ${25}^{\circ} \text{C}$, which is equal to

$\rho = \text{0.99705 g/mL}$

http://antoine.frostburg.edu/chem/senese/javascript/water-density.html

This means that the mass of water will be

250.0color(red)(cancel(color(black)("mL"))) * "0.99705 g"/(1color(red)(cancel(color(black)("mL")))) = "249.26 g"

Since you're dealing with glucose, a non-volatile compound, the vapor pressure of the solution will depend solely on the mole fraction of water and of the vapor pressure of pure water.

Simply put, the vapor pressure of the solution will contain solely water vapor.

Now, use glucose and water's respective molar masses to determine how many moles of each you have

76.6color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180.156color(red)(cancel(color(black)("g")))) = "0.4252 moles glucose"

and

249.26color(red)(cancel(color(black)("g"))) * "1 mole water"/(18.015color(red)(cancel(color(black)("g")))) = "13.836 moles water"

The mole fraction of water will be the ratio between the number of moles of water and the total number of moles in the solution.

The total number of moles in the solution will be

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

${n}_{\text{total" = 0.4252 + 13.836 = "14.261 moles}}$

This means that the mole fraction of water will be

chi_"water" = (13.836color(red)(cancel(color(black)("moles"))))/(14.261color(red)(cancel(color(black)("moles")))) = 0.9702

The vapor pressure of the solution will thus be

${P}_{\text{sol" = chi_"water" * P_"water}}^{\circ}$

${P}_{\text{sol" = 0.9702 * "23.8 torr" = "23.091 torr}}$

Rounded to three sig figs, the answer will be

P_"sol" = color(green)("23.1 torr")