Consider a 9% by mass aqueous ethyl alcohol solution.

a.) Calculate the molality of this solution_____m

b.) Given that ethyl alcohol is a molecular solute, calculate the freezing point of this solution. ________ degrees Celsius.

c.) Consider the boiling point of each solution. which one would raise the boiling temperature? _________ lower the boiling temperature? _________ explain.

a 9% by mass means 9g ethanol in 100 g solution or 9g ethanol/(9g ethanol + 91g H2O). m = mols/kg solvent

convert 9g ethanol to mols. mol = grams/molar mass. Then m = mols/0.091 = ?

b
delta T = Kf*m

c
You don't have but one solution here; what are you comparing?

To calculate the molality of the solution, we need to know the mass of ethyl alcohol in the solution and the mass of the solvent (water).

a.) To calculate the mass of ethyl alcohol in the solution, we multiply the mass percent (9%) by the total mass of the solution. Let's assume we have a 100g solution, then the mass of ethyl alcohol is 9g (100g x 0.09).

Now, let's calculate the molality. Molality (m) is defined as the moles of solute per kilogram of solvent. Since we know the mass of ethyl alcohol is 9g, we need to convert it to moles using its molar mass.

The molar mass of ethyl alcohol (C2H5OH) is approximately 46.07 g/mol. So, the number of moles of ethyl alcohol is 9g / 46.07g/mol = 0.195 moles.

Now, we need to calculate the mass of the solvent (water). Assuming the total solution mass is 100g and the mass of ethyl alcohol is 9g, the mass of water is 100g - 9g = 91g.

Finally, we can calculate the molality:
molality (m) = moles of solute / mass of solvent in kg
= 0.195 moles / (91g / 1000g/kg)
≈ 2.14 m

b.) To calculate the freezing point of the solution, we need to use the formula for the freezing point depression:
ΔT = Kf * m

In this case, ethyl alcohol is a molecular solute, so we can use the Van 't Hoff factor (i) of 1. The freezing point depression constant (Kf) for water is 1.86 °C/m. And we already calculated the molality (m) as 2.14 m.

ΔT = 1.86 °C/m * 2.14 m
≈ 3.98 °C

Since the freezing point of pure water is 0 °C, the freezing point of the ethyl alcohol solution can be calculated by subtracting the depression (3.98 °C) from 0 °C.

Freezing point = 0 °C - 3.98 °C
≈ -3.98 °C

c.) When we add a solute to a solvent, it affects the boiling point of the solution. The boiling point elevation is directly proportional to the molality of the solution.

Therefore, the solution with a higher molality will raise the boiling temperature, and the solution with a lower molality will lower the boiling temperature.

In this case, since we know the molality of the solution is 2.14 m, any solution with a higher molality than 2.14 m will raise the boiling temperature, and any solution with a lower molality will lower the boiling temperature.