Question

The vapor pressure of pure water at $25$ degrees Celsius is $23.8$ torr. What is the vapor pressure of a solution prepared by dissolving $18.0g$ of glucose (molecular weight = $180.0g/mol$) in $95.0g$ of water?

Answer 1

$23.4$ torr

Explanation:

For solutions that contain non-volatile solutes, the vapor pressure of the solution can be determined by using the mole fraction of the solvent, and the vapor pressure of the pure solvent at the same temperature.
$P_{sol}=X_{solvent}\cdot P_{solvent}^0$
where, $P_{sol}$ is the vapor pressure of the solution
$X_{solvent}$ is the mole fraction of the solvent
$P_{solvent}^0$ is the vapor pressure of the pure solvent.

In this case, you know that the vapor pressure of pure water at ${25}^{0}C$ is equal to $23.8$ torr. This means that all you have to do is determine the mole fraction of water in the solution.

As you know, a mole fraction is defined as the number of moles of a component of a solution divided by the total number of moles present in that solution.
Use glucose and water's respective molar masses to determine how many moles of each you have
$18.0g\cdot \frac{1moleofglucose}{180.0g}=0.100$ moles of glucose.
and
$95.0g\cdot \frac{1moleofwater}{18.015g}=5.273$ moles water

The total number of moles present in the solution will be
$n_{total}=n_{glucose}+n_{water}$
$n_{total}=0.100+5.273=5.373$ moles.

This means that the mole fraction of water will be
$X_{water}=\frac{5.272moles}{5.373moles}=0.9814$

Finally, the vapor pressure of the solution will be
$P_{sol}=0.9814\cdot 23.8torr=23.4torr$