A solution is made by dissolving 0.584 mol of nonelectrolyte solute in 883 g of benzene. Calculate the freezing point and boiling point of the solution.

Tf=?
Tb=?

I got Tf as 3.39 degrees Celsius
I got Tb as 1.67 degrees Celsius

It marked me wrong and said "You have given the change in the freezing and boiling points. Identify the normal freezing and boiling points of benzene, then add or subtract the change."

Can you please explain and let me know the correct answers so I don't miss it again. Thanks.

The comment made by the teacher explains what to do. Look up the freezing and boiling points of pure benzene and apply the appropriate corrections.

You did not calculate the freezing and boiling points. You calculated the changes from the pure compound (benzene), and you did not say if the corrections were up or down.

For PURE benzene,
Tf = 5.51 C
Tb = 80.10 C

You calculated delta T which is in the formula delta T = Kf*m and delta T = Kb*m

The problem asked for freezing point and boiling point.
What is the normal freezing point of benzene. It's approximately 5.50 C (you should confirm that and use the number in your text/note table. Then 5.50-3.39 = ?(I didn't check your 3.39 since I don't have the Kf and Kb memorized for benzene.)

The normal boiling point for benzene is approximately 80 C (again, use the number in your text/notes) so the new boiling point will be 80 + 1.67 (and I didn't confirm the 1.l67 for the same reasons I cited above.).
But this should take care of it I think.

So do I subtract 3.39- 5.51

and 1.67- 80.10? or are those the answers?

I'm still confused.

I got it! Thanks for the help...both of you :)

No. The normal f.p. for benzene is about 5.5. Your delta T is 3.39. That 3.39 is how much the f.p. has been lowered so the new f.p. is 5.l5-3.39 = ?

The 1.67 is delta T for the boiling point. That tells you how much the b.p. has been raised. So the new b.p. is the old one (about 80) + 1.67 = ?
Remember what happens. When a non-volatile solute is added to a volatile solvent three things happen. The freezing point is lowered, the boiling point is raised, and the vapor pressure is lowered. The extent to which these three things happen depends SOLELY upon the number of dissolved particles (;i.e., not the solute or the solvent).

Ohhh yeah i got it already. Thanks.

To calculate the freezing point and boiling point of a solution, you need to consider the molality of the solute and the specific colligative properties of the solvent. In this case, we have a nonelectrolyte solute dissolved in benzene as the solvent.

The freezing point depression and boiling point elevation equations are as follows:

For the freezing point depression (ΔTf):
ΔTf = Kf × m

For the boiling point elevation (ΔTb):
ΔTb = Kb × m

Where Kf is the freezing point depression constant and Kb is the boiling point elevation constant specific to the solvent, and m is the molality of the solution.

Since benzene is the solvent, we need to use the freezing point depression and boiling point elevation constants for benzene. The freezing point depression constant (Kf) for benzene is 5.12 °C/m, and the boiling point elevation constant (Kb) is 2.53 °C/m.

First, we need to calculate the molality (m) of the solution. Molality is the moles of solute per kilogram of solvent. In this case, we have 883 g of benzene, which is equivalent to 0.883 kg.

m = moles of solute / mass of solvent in kg

m = 0.584 mol / 0.883 kg
m = 0.661 mol/kg

Now, let's calculate the freezing point depression (ΔTf) and boiling point elevation (ΔTb) using the equations above:

ΔTf = Kf × m
ΔTf = 5.12 °C/m × 0.661 mol/kg
ΔTf = 3.38192 °C

To obtain the actual freezing point, we need to subtract ΔTf from the normal freezing point of benzene, which is 5.5 °C.

Tf = Normal freezing point - ΔTf
Tf = 5.5 °C - 3.38192 °C
Tf = 2.11808 °C

Thus, the freezing point of the solution is approximately 2.12 °C.

Similarly, let's calculate the boiling point elevation:

ΔTb = Kb × m
ΔTb = 2.53 °C/m × 0.661 mol/kg
ΔTb = 1.67293 °C

To obtain the actual boiling point, we need to add ΔTb to the normal boiling point of benzene, which is 80.1 °C.

Tb = Normal boiling point + ΔTb
Tb = 80.1 °C + 1.67293 °C
Tb = 81.77293 °C

Thus, the boiling point of the solution is approximately 81.77 °C.

Please note that these calculations are approximate due to rounding.