Chemistry Freezing Point questions?

Question

1) How much anthracene (MW= 178.23 g/mol) would need to be added to 100.0g of cyclohexane (Kfp= 20.4 degrees C/m) to depress the melting point by 5.50 degrees C? Anthracene is a non-electrolyte.

2) A solution with 8.35g of a non-electrolyte solute in 50.0g naphthalene (Kfp= 6.90 degrees C/m) freezes at 76.5 degrees C. Pure naphthalene freezes at 81.0 degrees C. What is the molar mass of the solute?

3) If 0.725g of calcium nitrate was added to 4.25g of water (Kfp= 1.86 degrees C/m), what would you expect the freezing point of that solution to be?

If you could show how you've done it, that would be incredibly helpful. I'm stuck and genuinely don't know how to do this.

Thank you in advance !!

hcbiochem · Accepted Answer

In all of these problems, the fundamental equation is:

Delta Tf = i Kfp m
where
Delta Tf is the freezing point depression in degrees C
i is the number of ions formed (if the substance is an electrolyte)
Kfp is the freezing point depression constant
m is the molality of the solution, which is calculated as moles of solute / kilogram solvent

1) 5.50 C = 20.4 C/m (m)
m = 0.2696 mol anthracene/kg solvent
Here, you have 0.1000 kg solvent, so,
moles solute = 0.2696 mol/kg X 0.1000 kg = 0.02696 mol
mass anthracene = 0.2696 mol X 178.23 g/mol = 4.81 g anthracene

2) Delta Tf = Kfp m
(81.0 - 76.5) = 6.90 C/m (m)
m = 0.652 mol/kg

The actual solution was made as 8.35 g/0.0500 kg = 167 g/kg

Dividing those two values gives 167 g/kg / 0.652 mol/kg = 256 g/mol

3) moles Ca(NO3)2 = 0.725 g / 164.1 g/mol = 4.42X10^-3 mol
Ca(NO3)2 forms 3 ions in solution, so
Delta Tf = 3 (1.86 C/m) (4.42X10^-3 mol / 0.00425 kg) = 5.80
Because the normal freezing point of water is 0 C, this solution would freeze at -5.80 C.

Hope that helps....

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Chemistry Freezing Point questions?

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