# Chemistry Lab equations

1. Jul 15, 2010

### Erica23

Chemistry Lab equations :(

1. The problem statement, all variables and given/known data[/b]
Here is the question:
-A solution was prepared by dissolving 0.46 kg of ethylene glycol (C2H6O2) in 2.46 kg of water.
A.) calculate the molal solution
B.) calculate the expected freezing point of the solution

2. Relevant equations

m= (moles of solute)/(kg of solvent)

change (delta) Tf =Tf - T'f =Kf(m)

3. The attempt at a solution

Question a:
atomic weights of ethylene glycol (C2H6O2) (12 x 2) + (1 x 6) + (16 x 2) = approx 62

moles of (C2H6O2): 46 g x (1 mol)/62 g = .742 mol

m= .742 mol/2.46 kg water = .3 mol

**** the answer to this problem is supposed to be 3 m... so how do I only get .3? What am I doing wrong? I have e-mailed and asked my teacher, but she is too busy to give me a straight answer that I understand and won't explain the steps.

Question b:
change (delta) Tf =Tf - T'f =Kf(m)

Tf= 0 degrees C (since the initial solvent is water and water's freezing point is 0 degrees C)
m= 0.742 from the previous equation... if it is correct.

The problem I am having with this one is that I don't know where to get the additional information for the remaining constants. How do I know what Kf is? Any help would be greatly appreciated as this is due tomorrow, and my teacher has STILL not e-mailed me back :( Thanks!

2. Jul 15, 2010

### Bohrok

Re: Chemistry Lab equations :(

You didn't convert 0.46 kg to grams correctly, off by one decimal place.

For part B, m is the molality of the solute, not moles of the solute.
Lots of common solvents have kf values which are freezing point depression constants. kf of water is 1.86°C/m. Plug that into the equation, and remember that that gives you only the freezing point depression, there's one more step to find the actual freezing point.

3. Jul 15, 2010

### Erica23

Re: Chemistry Lab equations :(

Thank you! You are my hero :)

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