[solomon_pup]

Shalom and thanks again for the help to everyone on this site.  I have been having a bit of trouble with this problem and I was wondering if someone could tell me if I have found the right approach to this problem:

The question:
A radiator is to be protected against freezing to -20F.  Calculate the volume of ethylene glycol C₂H₄(OH)₂ needed per liter of water.  Density of glycol is 1.11 g/mL.

My answer :  1.17g Glycol/L H₂O

My approach: 
First I converted F to C and came out with -29C. 
Then I figured Delta T to be 29 because the normal Freezing point of water is 0C. 
Then I set Delta T (29) equal to (1.56)X.  (I used 1.56 because it is the K_f for H₂0.  I came up with 18.9, which I assumed to be my molality because Delta T = K_fXMolality.
Then I went backwards.  I assumed 1000 mL of glycol.  I said 1.11=x/1000 because D of glycol is 1.11 g/mL.  I came up with 1,110.  I then divided 1,110 by 62 to find the moles.  I found that to be 17.9 moles.  .  I then set moles (17.9)/X = 18.9 because the moles over kg of solvent unknown equals molality.  I came up with .95 kg. 
Then I set 17.9/.95 = X/.001   (because D of Water is 1g/mL and I'm trying to find for one liter.)  I came up with .02 moles, then multiplied that by 62 g because 62 g  = 1 mole glycol and came up with 1.17 g glycol/ L H₂O

Does this sound like the right approach?  Unfortunately I had to miss class due to an emergency so I didn't get to hear the lecture.  This is the most I made out of it.  If I am wrong I would appreciate any help in the situation. I fear I am wrong in assuming 1000 mL, but I didn't think that it would effect it because I figured it would come down to ratio.  Thank you so much!!

-JO

[Albert]

Calculate the volume of ethylene glycol C₂H₄(OH)₂ needed per liter of water.

My answer :  1.17g Glycol/L H₂O

Why is your answer in g/L? I think it should be expressed in mL/L or L/L.

[solomon_pup]

you're right I see why it should be g/mL (I just went and changed that) but I think it must be in grams, that's how I got 1.17 was by multiplying by 62 because they want they mass not the volume.  They want mass of glycol per L of water, so I need to change my answer so it matches Liters

[solomon_pup]

oh wait they do want volume per liter.  Damn

[Albert]

I used 1.56 because it is the K_f for H₂0.  I came up with 18.9, which I assumed to be my molality because Delta T = K_fXMolality.

1.56? Where did you find it? I'd use 1.858 (°C/m) or something like that.

[firzzy87]

When doing this topics i always can't differ between molality and molarity ... The best part is its formula is so easy to remember ... especially relating Kb and Kf

[sdekivit]

molarity = mol solute / L solvent

molality = mol solute / kg solvent

[firzzy87]

molarity = mol solute / L solvent

molality = mol solute / kg solvent

so if we convert from kg to L , is it effect the molarity and molality ...