the freezing point of ethanol (C2H5OH) is -114.6C. the molal freezing point depression constant for ethanol is 2.00C/m. what is the freezing point (C) of a solution prepared by dissolving 50.0 g of glycerin (C3H8O3, A nonelectrolyte) in 200 g of ethanol.

any help would be appreciated im very confused.

To find the freezing point of the solution, we can use the equation:

ΔT = Kf * m

Where ΔT is the freezing point depression, Kf is the molal freezing point depression constant, and m is the molality of the solution.

First, let's calculate the molality of the solution.

Molality (m) is defined as the number of moles of solute (glycerin) divided by the mass of the solvent (ethanol) in kilograms.

Step 1: Calculate the moles of glycerin (C3H8O3) using its molar mass.

Molar mass of C3H8O3 = (3 * atomic mass of C) + (8 * atomic mass of H) + (3 * atomic mass of O)
= (3 * 12.01 g/mol) + (8 * 1.01 g/mol) + (3 * 16.00 g/mol)
= 92.09 g/mol

Moles of glycerin = mass of glycerin / molar mass
= 50.0 g / 92.09 g/mol
= 0.5439 mol

Step 2: Convert the mass of ethanol to kilograms.

Mass of ethanol = 200 g
= 200 g / 1000 (convert to kg)
= 0.2 kg

Step 3: Calculate the molality.

Molality (m) = moles of glycerin / mass of ethanol (in kg)
= 0.5439 mol / 0.2 kg
= 2.72 mol/kg

Now, we can calculate the freezing point depression (ΔT).

ΔT = Kf * m
ΔT = 2.00°C/m * 2.72 mol/kg
ΔT = 5.44°C

Finally, to find the freezing point of the solution, subtract the freezing point depression (ΔT) from the freezing point of the pure solvent (ethanol).

Freezing point = Freezing point of pure solvent - ΔT
= -114.6°C - 5.44°C
= -120.04°C

Therefore, the freezing point of the solution prepared by dissolving 50.0 g of glycerin in 200 g of ethanol is approximately -120.04°C.