Question

Beaker \(\mathrm{A}\) has \(1.00 \mathrm{~mol}\) of chloroform, \(\mathrm{CHCl}_{3}\), at \(27^{\circ} \mathrm{C}\). Beaker \(\mathrm{B}\) has \(1.00 \mathrm{~mol}\) of carbon tetrachloride, \(\mathrm{CCl}_{4}\), also at \(27^{\circ} \mathrm{C}\). Equal masses of a nonvolatile, nonreactive solute are added to both beakers. In answering the questions below, the following data may be helpful. $$ \begin{array}{lll} \hline & \mathrm{CHCl}_{3}(\mathrm{~A}) & \mathrm{cCl}_{4}(\mathrm{~B}) \\ \hline \text { Vapor pressure at } 27^{\circ} \mathrm{C} & 0.276 \text { atm } & 0.164 \text { atm } \\ \text { Boiling point } & 61.26^{\circ} \mathrm{C} & 76.5^{\circ} \mathrm{C} \\\ k_{\mathrm{b}}\left({ }^{\circ} \mathrm{C} / \mathrm{m}\right) & 3.63 & 5.03 \\\ \hline \end{array} $$ Write \(<,>,=\), or more information needed in the blanks provided. (a) Vapor pressure of solvent over beaker B of solvent over beaker \(\mathrm{A}\). (b) Boiling point of solution in beaker \(A\) boiling point of solution in beaker B. (c) Vapor pressure of pure \(\mathrm{CHCl}_{3} \longrightarrow\) vapor pressure of solvent over beaker \(\mathrm{A}\). (d) Vapor pressure lowering of solvent in beaker A pressure lowering of solvent in beaker B. (e) Mole fraction of solute in beaker \(\mathrm{A}\) mole fraction of solute in beaker \(\mathrm{B}\).

Short answer

Question: Compare the vapor pressure of solvent over beaker B to that of solvent over beaker A after a nonvolatile, nonreactive solute is added. Answer: Vapor pressure of solvent over beaker B < Vapor pressure of solvent over beaker A.

Step-by-step solution

Understand Raoult's Law for vapor pressure

Raoult's Law states that the vapor pressure of a solvent in a solution is proportional to its mole fraction: $$P_{A} = x_{A}P^{0}_{A}$$ Where \(P_{A}\) is the vapor pressure of solvent A in the solution, \(x_{A}\) is the mole fraction of A, and \(P^{0}_{A}\) is the vapor pressure of pure A.

Compare vapor pressures

Initially, Beaker A has a vapor pressure of 0.276 atm and Beaker B has a vapor pressure of 0.164 atm. Adding a nonvolatile solute will decrease the mole fraction of the solvent in both beakers and thus decrease the vapor pressure. Since the vapor pressure of Beaker A was initially higher, the vapor pressure of the solvent over Beaker A will remain higher than that of the solvent over Beaker B. Therefore: $$\text{Vapor pressure of solvent over beaker B} < \text{Vapor pressure of solvent over beaker A}$$ #b) Boiling point of solution in beaker A boiling point of solution in beaker B#

Understand the boiling point elevation equation

The boiling point elevation equation is as follows: $$\Delta T_{b} = k_{b} \cdot m$$ Where \(\Delta T_{b}\) is the change in boiling point, \(k_{b}\) is the molal boiling point elevation constant, and \(m\) is the molality of the solute.

Compare the boiling points of the two solutions

Since both beakers have equal masses of the nonvolatile, nonreactive solute added, their molalities are similar. Therefore, the boiling points of the solutions are proportional to their \(k_b\) values. Beaker A has a lower \(k_{b}\) of 3.63 compared to Beaker B's \(k_{b}\) of 5.03. Thus, the boiling point elevation in Beaker A will be smaller than that in Beaker B. Since Beaker A has a lower initial boiling point (61.26°C) than Beaker B (76.5°C), we cannot directly determine whether the new boiling point of the solution in Beaker A would be higher or lower than that of the solution in Beaker B without more information on the molality of the solute. Hence, the answer is: $$\text{More information needed}$$ #c) Vapor pressure of pure CHCl3 vapor pressure of solvent over beaker A#

Compare the vapor pressures

The vapor pressure of pure CHCl3 is given as 0.276 atm. Adding a solute to the CHCl3 in Beaker A will decrease the mole fraction of CHCl3, which results in a reduced vapor pressure according to Raoult's Law. Thus, we can write: $$\text{Vapor pressure of pure CHCl3} > \text{Vapor pressure of solvent over beaker A}$$ #d) Vapor pressure lowering of solvent in beaker A pressure lowering of solvent in beaker B#

Understand the vapor pressure lowering concept

Vapor pressure lowering refers to the decrease in the vapor pressure of a solvent in a solution compared to its vapor pressure as a pure substance. This change is proportional to the mole fraction of the solute.

Compare the vapor pressure lowering

Since equal masses of the solute are added to both beakers, the mole fraction of the solute in both solutions will be similar. Thus, the vapor pressure lowering in both Beaker A and Beaker B will also be similar: $$\text{Vapor pressure lowering of solvent in beaker A} = \text{Vapor pressure lowering of solvent in beaker B}$$ #e) Mole fraction of solute in beaker A mole fraction of solute in beaker B#

Compare the mole fractions of solute

Since the same mass of solute is added to both beakers, and both beakers originally contain 1 mol of their respective components, the mole fractions of solute in each beaker will also be equal. Thus, we can write: $$\text{Mole fraction of solute in beaker A} = \text{Mole fraction of solute in beaker B}$$

Key concepts

Raoult's Law

The understanding of Raoult's Law is fundamental when it comes to vapor pressure in chemistry. It states that the vapor pressure of a solvent in an ideal solution is directly proportional to the mole fraction of the solvent in the solution. Mathematically, this is expressed as
\[ P_{A} = x_{A}P^{0}_{A} \]
where P_{A} is the vapor pressure of the solvent in the solution, x_{A} is the mole fraction of the solvent, and P^{0}_{A} is the vapor pressure of the pure solvent.

This equation implies that by adding a nonvolatile solute, which does not contribute to the vapor pressure, the mole fraction of the solvent decreases, leading to a lower overall vapor pressure of the solvent in the solution. In the context of our exercise from the textbook, when a nonvolatile solute is added to chloroform or carbon tetrachloride, the vapor pressure over each solution decreases in proportion to the mole fraction of the solvent that remains. This concept is critical when predicting the changes in physical properties that result from the formation of a solution.

Vapor Pressure Lowering

The concept of vapor pressure lowering is directly tied to Raoult's Law. It describes the reduction in vapor pressure experienced by a solvent when a nonvolatile solute is dissolved into it. This is an important colligative property, meaning it depends on the number of solute particles in the solution, rather than their identity.

The reduction in vapor pressure is a result of fewer solvent molecules being present at the surface of the liquid, which decreases the rate at which solvent molecules escape into the vapor phase. This phenomenon can be quantitatively analyzed by calculating the difference between the vapor pressure of the pure solvent and the vapor pressure of the solvent in the solution. As shown in the step by step solution for the textbook exercise, when equal amounts of a nonvolatile solute are introduced into two solvents, the vapor pressure lowering effect should be the same for both, assuming the solutes do not interact differently with the respective solvents.

In practice, this means that if Beaker A and Beaker B initially contain equal molar amounts of two different solvents and the same mass of solute is added to both, the resulting decrease in vapor pressure should be comparable for both solutions.

Boiling Point Elevation

Another colligative property of solutions is boiling point elevation, which reflects how the presence of a solute affects the temperature at which the solvent boils. Boiling point elevation occurs because the addition of a solute lowers the vapor pressure of the solution; this means more heat (energy) is required to increase the vapor pressure to atmospheric pressure, thus raising the boiling point.

The boiling point elevation can be explained quantitatively by the equation:
\[ \Delta T_{b} = k_{b} \cdot m \]
where \Delta T_{b} is the boiling point elevation, k_{b} is the molal boiling point elevation constant specific to each solvent, and m is the molality of the solute in the solution. We see in the provided exercise that we cannot definitively say which solution has a higher boiling point after adding the solute, because although we know the constants k_{b} and the original boiling points of the pure solvents, the exercise does not give us the molality of the solute in each solution.

As a rule, the greater the number of solute particles in a solvent, the more pronounced the boiling point elevation will be. This is a crucial factor when determining boiling points in mixtures, provided the solute does not undergo any reaction or volatilization.