A solution prepared by dissolving 3.00 grams of ascorbic acid (vitamin C, C6H8O6), in

50.0 grams of acetic acid has a freezing point that is depressed by �T = 1.33 oC below that of
pure acetic acid. What is the value of the molar freezing point-depression constant for acetic acid?

mols ascorbic acid = grams/molar mass

m = mols/kg solvent

Then 1.33 = Kf*m
Substitut for m and solve for Kf.

^ but what about the i?

I assume the problem expects you to use i = 1.

To find the value of the molar freezing point-depression constant (Kf) for acetic acid, we need to use the formula:

ΔT = Kf * m * i,

where:
- ΔT is the freezing point depression (given as 1.33°C),
- m is the molality of the solution (mol solute/kg solvent),
- i is the van't Hoff factor, which represents the number of particles the solute dissociates into in the solution. In this case, acetic acid does not dissociate significantly, so i = 1.

First, let's find the molality (m) of the solution:
Molality (m) = moles of solute / mass of solvent (in kg)

We are given that 3.00 grams of ascorbic acid (C6H8O6) is dissolved in 50.0 grams of acetic acid.

Step 1: Convert grams to moles.
To find the moles of ascorbic acid:
moles = mass / molar mass
molar mass of ascorbic acid (C6H8O6) = (6 * atomic mass of carbon) + (8 * atomic mass of hydrogen) + (6 * atomic mass of oxygen)
= (6 * 12.01 g/mol) + (8 * 1.01 g/mol) + (6 * 16.00 g/mol)

Step 2: Calculate the moles of acetic acid.
Since acetic acid doesn't dissociate significantly, we can assume its molar mass is equal to its molecular weight, which is around 60.052 g/mol.

Step 3: Calculate molality (m) in mol solute/kg solvent.
m = moles of ascorbic acid / (mass of acetic acid / 1000) (convert kg to g)

Step 4: Substitute the values into the formula to solve for Kf:
1.33°C = Kf * m * 1

Finally, solve for Kf by rearranging the equation:
Kf = 1.33°C / (m * i)

Substitute the value of m and i into the equation to find the value of Kf.