To understand the scenario you've presented, we need to delve into the concepts of solutions, freezing point depression, and vapor pressure. Let's break it down step by step.
Understanding the Solution
In your example, we have a solution made by dissolving cane sugar (sucrose) in water. Cane sugar is a non-volatile solute, meaning it does not evaporate into vapor like water does. When you dissolve sugar in water, it affects the physical properties of the solution, including its freezing point and vapor pressure.
Freezing Point Depression
When a solute like sugar is added to a solvent (water, in this case), the freezing point of the solution decreases compared to that of pure water. This phenomenon is known as freezing point depression. The more solute you add, the lower the freezing point becomes. In your case, the solution was cooled below its freezing point, causing some ice to form. This indicates that the concentration of the sugar solution was significant enough to lower the freezing point below 0°C.
Vapor Pressure of the Solution
The vapor pressure of a solution is the pressure exerted by the vapor in equilibrium with its liquid phase. When you mentioned that the remaining aqueous solution registered a vapor pressure of 700 mm Hg, this tells us about the effect of the dissolved sugar on the vapor pressure of the solution. In general, the presence of a solute decreases the vapor pressure of the solvent due to the solute particles occupying space at the surface of the liquid, which reduces the number of solvent molecules that can escape into the vapor phase.
Calculating the Effect of Sugar on Vapor Pressure
To analyze the vapor pressure further, we can use Raoult's Law, which states that the vapor pressure of a solvent in a solution (P_solution) is equal to the vapor pressure of the pure solvent (P°_solvent) multiplied by the mole fraction of the solvent (X_solvent) in the solution:
- P_solution = P°_solvent × X_solvent
In this case, if we know the vapor pressure of pure water at the same temperature (which is approximately 760 mm Hg at 0°C), we can rearrange the equation to find the mole fraction of water in the solution:
- X_solvent = P_solution / P°_solvent
Substituting the values:
- X_solvent = 700 mm Hg / 760 mm Hg ≈ 0.921
Finding the Mole Fraction of Sugar
Since the mole fractions of all components in a solution must sum to 1, we can find the mole fraction of sugar (X_sugar):
- X_sugar = 1 - X_solvent = 1 - 0.921 ≈ 0.079
Implications of the Results
This calculation shows that approximately 7.9% of the solution's moles are sugar. This relatively low concentration is consistent with the fact that sugar is a non-volatile solute, which significantly alters the physical properties of the solution.
In summary, the cooling of the sugar solution below its freezing point resulted in ice formation, while the vapor pressure of the remaining solution was measured at 700 mm Hg. This scenario illustrates the principles of colligative properties, which depend on the number of solute particles in a solution rather than their identity. Understanding these concepts is crucial in fields such as chemistry and food science, where solutions play a vital role in various processes.
