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What is the freezing point of a solution of ethyl alcohol, that contains 26.3 g of the solute (C2H5OH), diss

Submitted by antgoblue on Sat, 05/03/2008 - 12:19

I did this problem and can't quite get the right answer. Could someone point me in the right direction!

What is the freezing point of a solution of ethyl alcohol, that contains 26.3 g of the solute (C2H5OH), dissolved in 760 g of water?

!. Find molality

26.3g x 1mol/46g =0..57 mol

.57mol/.76 kg= 0.75 m

Ok, now i'm confused and don't know what to do next!!
I also need some hel with a boling point question on a topic below!!

Thanks for the guidance!!


Once you know the molality you plug into the equation:
       
             change in freezing pt = Kf  x molality  (where Kf is the freezing point depression constant for a particular solvent - Kf for water is 1.86C/molal)

After finding the change in temperature, you subtract it from the normal freezing point of the solvent.

Boiling point problems are done in exactly the same way.
1) Find moles of solute
2) Find molality of solution
3) Find change in boiling point:
              change in boiling pt = Kb  x molality  (where Kb is the boiling point elevation constant for a particular solvent - Kb for water is .52C/molal)
4) Add the change in boiling point to the normal boiling point of the solvent.

NOTE: Don't forget the boiling point is raised and the freezing point is lowered by adding a solute.


Thank You Spock, thats exactly what I needed. I got it correct. Just one more question, what do you do to find the molecular mass like in this problem---

  The molal boiling point constant for ethyl alcohol is 1.22 oC/molal. Its boiling point is 78.4 oC. A solution of 17.3 g of a nonvolatile nonelectrolyte in 270 g of alcohol boils at 79.9 oC. What is the gram molecular mass of the solute?


You work backwards!

1) Figure out the change in boiling point by taking the difference between the new boiling point (79.9)and the boiling point of the pure solvent (78.4C)

2) Use the boiling point formula (change in boiling pt = Kb x molality).  Since you know the change in boiling point (from step 1) and the Kb (1.22C/molal), you can find the molality.

3)  Use the molality formula (molality = moles solute/kg solvent).  Since you know the molality (from step 2 above) and the mass of the solvent (.270 kg), you can find the number of moles of solute.

4) Divide the number of grams of solute (17.3g) by the moles of solute (from step 3 above) to get the gram molecular mass (the number of grams per mole).


Spock I did all that and can't get the correct answer!! What answer did you get??

I got
79.9-78.4=1.5 temp diff.
molality=change/kb =1.5/1.22= 1.23
moles=1.23/.270kg=4.5556
17.3g/4.556=3.7 g/mol
and my program says its incorect!!
Did I mess up somewhere?


I must be making a mistake somewhere because i can't get another one either. I think its the last step i'm confused on!


Your mistake is in step 3:

3)  Use the molality formula (molality = moles solute/kg solvent).  Since you know the molality (from step 2 above) and the mass of the solvent (.270 kg), you can find the number of moles of solute.

You wrote:
moles=1.23/.270kg=4.5556
You either plugged in incorrectly or solved the equation incorrectly:
               
                          molality = moles solute / kg solvent

You have found the molality and are looling for the moles of solute:

                            1.22 molal =  x moles / .270 kg

Solving for x yields:
                          x moles = 1.22 molal  x  .270 kg

After finding the correct number of moles you should be able to get the correct answer.


Sweet Spock I got two more right now only have 2 left! I'm confused on how to the last 2 though. Heres one of them:

How many grams of ethylene glycol, C2H4(OH)2, must be added to 800 g of water to yield a solution that will freeze at -8.55 oC?


still can't figure out how to starrt!!!!


Almost exactly the same as the last problem except for the fact that you are using freezing point rather than boiling point and the last step:
You work backwards!

1) Figure out the change in freezing point by taking the difference between the new boiling point (-8.55C)and the boiling point of the pure solvent (0C)

2) Use the freezing point formula (change in freezing pt = Kf x molality).  Since you know the change in freezing point (from step 1) and the Kf (1.86C/molal), you can find the molality.

3)  Use the molality formula (molality = moles solute/kg solvent).  Since you know the molality (from step 2 above) and the mass of the solvent (.270 kg), you can find the number of moles of solute.

4) Convert from moles of solute (from the previous step) to grams by multiplying the moles by the molecular mass (from the formula of ethylene glycol).


Alright Spock! I got that one correct bt can't get this last one------

  A mass of 59 g of an unknown nonelectrolyte is dissolved in 439 g naphthalene. The nonelectrolyte lowers naphthalene's freezing point by 8.5 oC. What is the molecular mass of the unknown substance? (Kf = 6.8 oC/molal)

I did--
1. 0-8.5=8.5 temp. change
2. 8.5/6.8=1.25 m
3. 1.25 m x .439 kg =2.847 moles of solute
4. 2.847 moles/1mol x 128.17052 g=364.950 g/mol and this is incorrect. Where did I mess up this time around???