# Question 53001

Feb 4, 2016

${P}_{\text{sol" = "23.46 torr}}$

#### Explanation:

Notice that the problem provides you with the volume of water and its density at ${25}^{\circ} \text{C}$.

This should automatically tell you that you can find its mass by using the definition of density

$\textcolor{b l u e}{\text{density" = "mass"/"unit of volume}}$

Water is said to have a density of $\text{0.9971 g/mL}$ at that temperature, which means that every milliliters of water will have a mass of $\text{0.9971 g}$.

If that's the case, then $\text{643.5 mL}$ will have a mass of

643.5 color(red)(cancel(color(black)("mL"))) * overbrace("0.9971 g"/(1color(red)(cancel(color(black)("mL")))))^(color(purple)("density")) = "641.634 g"

Now, the vapor pressure of the solution will depend on the mole fraction of the solvent in the solution, ${\chi}_{\text{solvent}}$, and on the vapor pressure of the pure solvent, ${P}_{\text{solvent}}^{\circ}$ - think Raoult's Law.

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

In order to find the mole fraction of the solvent, you need to know the number of moles of solvent and the total number of moles present in solution.

Since you have the mass of water used to make this solution, use water's molar mass to help you find the number of moles it contains

641.634 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "35.617 moles H"_2"O"

To find the total number of moles present in solution, find the number of moles of solute first

158.0 color(red)(cancel(color(black)("g"))) * "1 mole sucrose"/(342.3 color(red)(cancel(color(black)("g")))) = "0.4616 moles sucrose"

The solution will thus contain a total number of

${n}_{\text{total" = "35.617 moles" + "0.4616 moles" = "36.079 moles}}$

The mole fraction of water will be

chi_"water" = (35.617color(red)(cancel(color(black)("moles"))))/(36.079color(red)(cancel(color(black)("moles")))) = 0.9872

Therefore, the vapor pressure of the solution will be

${P}_{\text{sol" = 0.9872 * "23.76 torr}}$

P_"sol" = color(green)("23.46 torr") -># rounded to four sig figs