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.

i tried to solve it on my own but im confused about the last part, this i what i have so far..

moles of glycerin= 50.0/92.10=0.54
molality=0.54/0.200=2.7

im a little confused about the numbers im supposed to use for the final answer...

To solve this problem, you need to use the equation for freezing point depression:

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, calculate the molality of the solution. You correctly calculated the moles of glycerin (0.54 mol), but there is a mistake in the calculation of molality. Molality is defined as moles of solute divided by kilograms of solvent. In this case, the solvent is ethanol, so you need to convert the mass of ethanol (200 g) to kilograms by dividing it by 1000:

molality = 0.54 mol / (200 g / 1000) = 0.54 mol / 0.2 kg = 2.7 mol/kg

Now that you have the molality of the solution, you can use the freezing point depression constant (2.00°C/m) to calculate the freezing point depression:

ΔT = Kf * m = 2.00°C/m * 2.7 mol/kg = 5.4°C

The freezing point depression (ΔT) is the difference between the freezing point of the pure solvent and the freezing point of the solution. Since the freezing point of ethanol is -114.6°C, the freezing point of the solution can be calculated by subtracting the freezing point depression from the freezing point of the pure solvent:

Freezing point of solution = -114.6°C - 5.4°C = -120.0°C

Therefore, the freezing point of the solution prepared by dissolving 50.0 g of glycerin in 200 g of ethanol is -120.0°C.