Question

An aqueous solution containing glucose has a vapor pressure of 19.6 torr at \(25^{\circ} \mathrm{C}\). What would be the vapor pressure of this solution at \(45^{\circ} \mathrm{C}\) ? The vapor pressure of pure water is 23.8 torr at \(25^{\circ} \mathrm{C}\) and 71.9 torr at \(45^{\circ} \mathrm{C} .\) If the glucose in the solution were substituted with an equivalent amount (moles) of \(\mathrm{NaCl}\), what would be the vapor pressure at \(45^{\circ} \mathrm{C}\) ?

Short answer

The vapor pressure of the solution containing glucose at \(45^{\circ} \mathrm{C}\) is \(59.2\, \text{torr}\). When glucose is substituted with an equivalent amount of \(\mathrm{NaCl}\), the vapor pressure of the solution at \(45^{\circ} \mathrm{C}\) is \(29.6\, \text{torr}\).

Step-by-step solution

Find the mole fraction of glucose del

Using the Raoult's Law for a non-volatile solute, we have: \[P_\text{solution} = P_\text{solvent} \cdot X_{\mathrm{A}}, \] where \(P_\text{solution}\) is the vapor pressure of the solution containing glucose, \(P_\text{solvent}\) is the vapor pressure of the pure solvent (water) and \(X_{\mathrm{A}}\) is the mole fraction of water (\(\mathrm{A}\)) in the solution. We will first calculate the mole fraction of glucose (\(\mathrm{A}\)) at \(25^{\circ} \mathrm{C}\). Given \(P_\text{solution} = 19.6\, \text{torr}\) and \(P_\text{solvent} = 23.8\, \text{torr}\), we can solve for \(X_{\mathrm{A}}\): \[X_\mathrm{A} = \dfrac{P_\text{solution}}{P_\text{solvent}} \]

Calculate the mole fraction of glucose

Plugging in the values of \(P_\text{solution}\) and \(P_\text{solvent}\): \[X_\mathrm{A} = \dfrac{19.6}{23.8} = 0.8235\] Now that we have the mole fraction of glucose, we can find the vapor pressure of the solution at \(45^{\circ} \mathrm{C}\).

Find the vapor pressure of the solution at 45°C

We can use the same Raoult's law formula with the obtained mole fraction(isothermical process) and vapor pressure of water at \(45^{\circ} \mathrm{C}\), which is given as \(71.9\, \text{torr}\): \[P_\text{solution} = P_\text{solvent} \cdot X_\mathrm{A}\] \[P_\text{solution} = 71.9\, \text{torr} \times 0.8235 \] \[P_\text{solution} = 59.2\, \text{torr}\] So, the vapor pressure of the solution containing glucose at \(45^{\circ} \mathrm{C}\) is \(59.2\, \text{torr}\). Now we will calculate the vapor pressure if glucose is substituted with an equivalent amount (moles) of \(\mathrm{NaCl}\).

Find the vapor pressure of the NaCl solution

When we substitute glucose with an equivalent amount of \(\mathrm{NaCl},\) the mole fraction of water will remain the same due to adding the same number of moles. However, as \(\mathrm{NaCl}\) dissociates into 2 ions in the solution, the one mole of glucose will be replaced by two moles of ions (one of \(\mathrm{Na}^+\) and one of \(\mathrm{Cl}^-\), so the effect( collusion ) on the vapor pressure will be doubled compared to glucose, which does not dissociate. Therefore, to find the mole fraction of water due to the substitution, we can multiply the obtained mole fraction by 0.5: \[X'_\mathrm{A} = 0.8235 \times 0.5 = 0.4118\] Now we can find the vapor pressure of the \(\mathrm{NaCl}\) solution using Raoult's law: \[P'_\text{solution} = P_\text{solvent} \cdot X'_\mathrm{A}\] \[P'_\text{solution} = 71.9\, \text{torr} \times 0.4118\] \[P'_\text{solution} = 29.6\, \text{torr}\] Therefore, the vapor pressure of the solution containing \(\mathrm{NaCl}\) at \(45^{\circ} \mathrm{C}\) is \(29.6\, \text{torr}\).

Key concepts

Vapor Pressure

Vapor pressure is a fundamental concept in chemistry that refers to the pressure exerted by the vapor in equilibrium with its liquid at a given temperature. This concept is crucial in understanding how solutions behave. When a non-volatile solute, like glucose, is added to a solvent such as water, the solute reduces the surface area available for evaporation, effectively lowering the vapor pressure of the solvent.

• The original solution vapor pressure depends on the mole fraction of the solvent.
• Vapor pressure is higher at elevated temperatures due to increased kinetic energy of molecules.

Understanding vapor pressure is essential for predicting how substances will behave in mixtures and during evaporation.

Mole Fraction

The mole fraction is a way to express the concentration of a component in a mixture. It is defined as the number of moles of a particular component divided by the total number of moles in the mixture. In our context, Raoult's Law is used to calculate vapor pressure using mole fraction.

• For a solvent, the formula is given by: \(X_{\text{A}} = \frac{\text{moles of solvent}}{\text{total moles in solution}}\).
• It provides a dimensionless number between 0 and 1, indicating how much of the solution the component contributes.

Mole fraction is crucial because it helps in applying Raoult's Law, which predicts how much the vapor pressure will decrease when a solute is added.

Aqueous Solution

An aqueous solution refers to a solution in which water is the solvent. Water is often called the "universal solvent" due to its ability to dissolve a wide range of substances. In the context of this problem, glucose dissolves in water to form an aqueous solution.

• Aqueous solutions are characterized by the interactions between water molecules and solute particles.
• They show unique properties like boiling point elevation and freezing point depression.

Understanding aqueous solutions is vital for solving problems involving the behavior of compounds in water, such as how solutes affect vapor pressure in this exercise.

Ion Dissociation

Ion dissociation refers to the process where an ionic compound, such as \(\text{NaCl}\), dissolves in water and separates into its constituent ions. This phenomenon plays a key role in altering the properties of a solution. When considering ion dissociation in the context of Raoult's Law, it impacts the mole fraction and consequently the vapor pressure of the solution.

• \(\text{NaCl}\) dissociates into \(\text{Na}^+\) and \(\text{Cl}^-\), effectively doubling the effect on the vapor pressure compared to a molecular solute like glucose.
• Ion dissociation increases the number of solute species in the solution, which leads to a more significant decrease in vapor pressure.

Grasping ion dissociation helps in predicting how ionic compounds will behave differently from non-ionic compounds in an aqueous solution.