Why do 0.60 grams of acetic acid dissolve in 200 grams of benzene to form a solution that lowers the freezing point of benzene to 5.40°C?

Question

Naomi Aronson · Accepted Answer

The freezing point is lowered because adding the acetic acid disrupts the dynamic equilibrium of the freezing process.

There is a dynamic equilibrium at the freezing point of benzene. Molecules of benzene leave the surface of the solid at the same rate as they return.

Adding acetic acid to the system will disrupt the equilibrium. The acetic acid molecules dissolve in the benzene, but they do not pack easily into the array of molecules in the solid.

Also, there are fewer benzene molecules on the liquid side because acetic acid has replaced some of the benzene molecules. The total number of benzene molecules captured by the solid per second goes down, so the rate of freezing goes down.

The presence of the acetic acid does not change the rate of melting, so melting occurs faster than freezing.

To re-establish equilibrium, you must cool the acetic acid-benzene mixture to below the normal freezing point of benzene.

You can learn how to calculate the freezing point depression at

https://tutor.hix.ai

In your problem, #i# = 1, and #K_"f"# = 5.12 °C/m.

#m# = 0.60 g acetic acid × #(1"mol acetic acid")/(60.05"g acetic acid")# = 0.009 992 mol acetic acid (2 significant figures + 2 guard digits)

#ΔT_"f" = iK_"f"m# = 1 x 5.12 °C/m × 0.0.009 992 m = 0.051 12 °C

For benzene,

#T_"f"^°# = 5.5 °C

#T_"f" = T_"f"^° - ΔT_"f"# = 5.5 °C – 0.051 12 °C = "5.449 °C" = 5.4 °C (1 decimal place)

Hope this helps.

Kai Bowman · Answer

undefined The phenomenon described is an example of colligative properties, specifically freezing point depression. Acetic acid, being a solute, disrupts the orderly arrangement of benzene molecules, reducing the freezing point of the solution below that of pure benzene. The extent of freezing point depression is proportional to the molality of the solute particles in solution, hence the observed decrease in freezing point. The relationship between freezing point depression, solute concentration, and the freezing point depression constant (Kf) is given by the equation: ΔTf = i * Kf * m, where ΔTf is the change in freezing point, i is the van't Hoff factor (number of particles the solute breaks into in solution), Kf is the freezing point depression constant, and m is the molality of the solute.

Jose Ayala · Answer

undefined This phenomenon occurs due to the process of freezing point depression, which is a colligative property of solutions. The presence of a solute, in this case, acetic acid, lowers the freezing point of the solvent, benzene, compared to its pure form. This happens because the solute particles disrupt the orderly arrangement of solvent molecules, making it more difficult for the solvent to freeze.

The extent of freezing point depression is proportional to the concentration of the solute particles in the solution. In this case, the molality of the solution, which is the number of moles of solute per kilogram of solvent, determines the degree of freezing point depression. By calculating the molality of the acetic acid solution, one can determine its effect on the freezing point of benzene.

Using the formula for freezing point depression, which states that ΔTf = Kf * m, where ΔTf is the change in freezing point, Kf is the cryoscopic constant, and m is the molality of the solution, we can solve for the molality of the acetic acid solution. Given that the freezing point depression is 5.40°C and Kf for benzene is 5.12°C/m, we can plug in these values to find the molality:

5.40°C = 5.12°C/m * m

m = 5.40°C / 5.12°C/m ≈ 1.05 m

This molality value indicates that there is approximately 1.05 moles of acetic acid per kilogram of benzene solvent. Therefore, the presence of 0.60 grams of acetic acid in 200 grams of benzene results in a solution with a molality of 1.05 m, causing the observed freezing point depression of benzene to 5.40°C.