Calculate Molal Solution and Freezing Point | Chemistry Lab Equations

[Erica23]

Chemistry Lab equations :(

1. Homework Statement [/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

Homework Equations

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

change (delta) T_f =T_f - T'_f =K_f(m)

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) T_f =T_f - T'_f =K_f(m)

T_f= 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 K_f is? Any help would be greatly appreciated as this is due tomorrow, and my teacher has STILL not e-mailed me back :( Thanks!

 

[Bohrok]

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 k_f values which are freezing point depression constants. k_f 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.

 

[Erica23]

Thank you! You are my hero :)