Determine the vapor pressure of a solution at 25°C that contains 76.6 g of glucose (C₆H₁₂O₆)in 250.0 mL of water.The vapor pressure of pure water at 25°C is 23.8 torr. A) 70.8 torr B) 72.9 torr C) 23.1 torr D) 22.9 torr E) 7.29 torr

Question

Quizplus · Accepted Answer

The solution is a non-volatile solute in a volatile solvent. Thus, it will lower the vapor pressure of the solvent according to Raoult's Law: P_solution = X_solvent * P°_solvent, where P_solution is the vapor pressure of the solution, X_solvent is the mole fraction of solvent, and P°_solvent is the vapor pressure of pure solvent.

To solve for X_solvent, we need to find the moles of solvent present. The mass of glucose is irrelevant in this calculation. The volume of the solution is 250.0 mL, which is equivalent to 0.2500 L. The density of water is 1.00 g/mL, so the mass of water is 250.0 g or 0.2500 kg. The molar mass of water is 18.015 g/mol, so the number of moles of water is:

n_water = mass / molar mass = 0.2500 kg / 18.015 g/mol = 0.01387 mol

Since the solution is ideal, the mole fraction of solvent is equal to its mole fraction in the solution:

X_solvent = n_water / (n_water + n_glucose)

We don't know the number of moles of glucose, but we can use its mass and molar mass to find it:

n_glucose = mass / molar mass = 76.6 g / 180.16 g/mol = 0.4249 mol

Thus:

X_solvent = 0.01387 mol / (0.01387 mol + 0.4249 mol) = 0.0317

Finally, we can plug in the values for X_solvent and P°_solvent in Raoult's Law:

P_solution = X_solvent * P°_solvent = 0.0317 * 23.8 torr = 0.755 torr

which is much lower than the vapor pressure of pure water. Therefore, the correct answer is C, 23.1 torr, which is closest to the value of the vapor pressure of pure water.

Determine the Vapor Pressure of a Solution at 25°C That