Question

Calculate the vapor pressure of water over each of the following solutions of glycerol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3},\) at \(28^{\circ} \mathrm{C}\) (vapor pressure of pure water \(=28.35 \mathrm{~mm} \mathrm{Hg})\). Glycerol can be assumed to be nonvolatile. (a) percent glycerol by mass \(=30.6 \%\) (b) \(2.74 \mathrm{~m}\) (c) mole fraction of glycerol \(=0.188\)

Short answer

The vapor pressures of water over the glycerol solutions at 28°C are: (a) 26.08 mm Hg (b) 27.02 mm Hg (c) 23.01 mm Hg

Step-by-step solution

(a) Given percent glycerol by mass = 30.6%

First, calculate the mass of glycerol and water in 100g of the solution. Since the percent glycerol by mass is 30.6%, the mass of glycerol is 30.6g and the mass of water is 69.4g (100g - 30.6g). Next, convert the mass of glycerol and water to moles using their respective molar masses. Molar mass of glycerol (\(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\)) is \(3\times12.01 + 8\times1.01 + 3\times16.00=92.09 \mathrm{~g/mol}\), and molar mass of water is \(18.02 \mathrm{~g/mol}\). Moles of glycerol = \(\frac{30.6}{92.09}=0.332\) Moles of water = \(\frac{69.4}{18.02}=3.849\) Mole fraction of water = \(\frac{moles \, of \, water}{total \, moles}=\frac{3.849}{3.849+0.332}=0.920\) Now, apply Raoult's law to find the vapor pressure of water: Vapor pressure = mole fraction of water × vapor pressure of pure water = \(0.920 \times 28.35=26.08 \mathrm{~mm} \mathrm{Hg}\).

(b) Given glycerol concentration = 2.74 m

Here, molarity (\(\text{m}\)) is given by moles of solute per kilogram of solvent. We have: \(2.74 \mathrm{~m} = \frac{moles \, of \, glycerol}{kg \, of \, water}\). Let's assume 1 kg = 1000 g of water. Moles of glycerol = \(2.74 \times 1 = 2.74\) Moles of water = \(\frac{1000}{18.02}=55.49\) Mole fraction of water = \(\frac{moles \, of \, water}{total \, moles}=\frac{55.49}{55.49+2.74}=0.953\) Using Raoult's law: Vapor pressure = mole fraction of water × vapor pressure of pure water = \(0.953 \times 28.35=27.02 \mathrm{~mm} \mathrm{Hg}\).

(c) Given mole fraction of glycerol = 0.188

To find the mole fraction of water, subtract the given mole fraction of glycerol from 1: Mole fraction of water = \(1 - 0.188 = 0.812\) Using Raoult's law: Vapor pressure = mole fraction of water × vapor pressure of pure water = \(0.812 \times 28.35=23.01 \mathrm{~mm} \mathrm{Hg}\). In summary, the vapor pressures of water over the glycerol solutions at 28°C are: (a) 26.08 mm Hg (b) 27.02 mm Hg (c) 23.01 mm Hg

Key concepts

Raoult's Law

Raoult's law is a fundamental principle in chemistry that relates the vapor pressure of a solvent in a solution to the mole fraction of the solvent. This law posits that the partial vapor pressure of an ideal solution's component is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture. In symbolic terms, if we denote the vapor pressure of the pure solvent as P₀, the mole fraction of the solvent as X_solvent, and the vapor pressure of the solvent when in solution as P_solvent, Raoult's law can be represented as:
\[ P_\text{solvent} = X_\text{solvent} \times P_0 \]
The law assumes that the solution behaves ideally, meaning the interactions between different molecules are similar to those in pure substances. Additionally, it's important to note that Raoult's law applies to the solvent in a solution where the solute is nonvolatile, as seen in the textbook exercise where glycerol is nonvolatile. The steps provided in the solution use Raoult's law to calculate the vapor pressure of water over glycerol solutions by first determining the mole fraction of water and then using it to calculate the new vapor pressure.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture. It is defined as the ratio of the number of moles of a particular component to the total number of moles of all components in the mixture. The mole fraction is dimensionless and represents the proportion of a substance within a mixture.
\[ X_\text{component} = \frac{\text{moles of component}}{\text{total moles in mixture}} \]
In the context of the exercise, to find the mole fraction of water, one must calculate the number of moles of water and glycerol separately and then divide the moles of water by the total number of moles. Using this value in Raoult's law enables one to determine the vapor pressure of water above the solution. The exercise solution elaborates this calculation stepwise, showing the importance of the mole fraction in determining colligative properties such as vapor pressure.

Colligative Properties

Colligative properties are characteristics of solutions that depend only on the ratio of the number of solute particles to solvent molecules and not on the nature of the substances. These properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.

When a nonvolatile solute is added to a solvent, the vapor pressure of the solvent is lowered because the solute particles take up space at the surface, resulting in fewer solvent molecules evaporating, which is vividly demonstrated in the exercise using glycerol solutions. Each colligative property provides insight into the molecular nature of solutions and can be used to determine molar masses and the strength of interactions within the mixture. In our exercise, the vapor pressure of water above the glycerol solutions is a clear representation of the colligative property of vapor pressure lowering due to the presence of a nonvolatile solute.