Benzaldehyde (106.12 g/mol), also known as oil of almonds, is used in the manufacture of dyes and perfumes and in flavorings. What would be the freezing point of a solution prepared by dissolving 75.00 g of benzaldehyde in 850.0 g of ethanol? (For ethanol, K_f = 1.99°C/m; freezing point of pure ethanol = -117.3°C) A) -117.5°C B) -118.7°C C) -119.0°C D) -120.6°C E) -121.1°C

Question

Quizplus · Accepted Answer

First, we need to calculate how many moles of benzaldehyde are dissolved in the solution:

75.00 g benzaldehyde / 106.12 g/mol = 0.707 mol benzaldehyde

Then, we can use the formula:

ΔTf = i*Kf*m

where ΔTf is the change in freezing point, i is the van't Hoff factor (which is 1 for benzaldehyde and ethanol), Kf is the freezing point depression constant for ethanol (1.99°C/m), and m is the molality of the solution, which is:

molality = moles of solute / mass of solvent (in kg)

We know that the mass of the solvent is 850.0 g = 0.850 kg, so:

molality = 0.707 mol / 0.850 kg = 0.831 m

Now we can plug in the values in the formula:

ΔTf = 1 * 1.99°C/m * 0.831 m = 1.65569°C

Finally, we can calculate the freezing point of the solution:

Freezing point = -117.3°C - ΔTf = -117.3°C - 1.65569°C = -118.95569°C

Rounding to the nearest tenth of a degree, we get:

Freezing point = -119.0°C

Therefore, the answer is choice C (-119.0°C).