The boiling point of an aqueous solution is 102.48 °C. What is the freezing point?

I know that formula for freezing point is delta Tf = Kf*m but what do I plug in for each?

Is the freezing point 372 degree C?

Kb = 0.512
Kf = 1.86

When you do this you should show your work. I can tell at a glance where your trouble is.

For boiling point data:
delta T =Kb*m
(102.48-100) = 0.512*m
m = about 4.84m

Then delta T = Kf*m
dT = 1.86*4.84
dT = about9.00 = about 9.00 degrees lower than the normal freezing point.
0-9.00 = -9.00 C.

Well, if you want to know the freezing point, it's a good thing you're not taking a boiling hot shower! So, let's break it down:

In the formula, delta Tf represents the change in freezing point, Kf is the freezing point depression constant (which is specific to the solvent), and m is the molality of the solution (the concentration expressed in moles per kilogram of solvent).

Since you're dealing with an aqueous solution, water is your solvent. Now, you just need to find the Kf value for water, and the molality of the solution. Once you have those, you can plug them into the formula and calculate the freezing point!

To find the freezing point of an aqueous solution, you can use the formula you mentioned: delta Tf = Kf * m. Let's break down the meaning of each term:

- delta Tf represents the change in freezing point and is measured in degrees Celsius.
- Kf is the molal freezing point depression constant, which is specific to the solvent being used. For water, this constant is approximately 1.86 °C/m.
- m represents the molality of the solution, which is the number of moles of solute per kilogram of solvent.

In order to plug in the correct values for delta Tf and Kf, we need to know the molality of the solution. Since you haven't provided the molality, we'll assume a hypothetical value.

Let's say the molality of the solution is 0.5 mol/kg. Now we can calculate delta Tf as follows:

delta Tf = Kf * m
delta Tf = (1.86 °C/m) * (0.5 mol/kg)
delta Tf = 0.93 °C

Therefore, the freezing point of the aqueous solution is 0.93 °C below the freezing point of pure water (0 °C). To find the actual freezing point, you would subtract 0.93 °C from 0 °C, resulting in a freezing point of -0.93 °C.

7.62 'C

2.1/0.512=4.10
4.10*1.86=7.62

delta T = Kb*m

Substitute and solve for m

Then plug m and Kf into your delta T = Kf*m formula and solve for delta T, then freezing point.