V. Determine the Molar Mass of glycerin from the freezing point depression data.
Mass of glycerin used: 2.84g
Freezing point of solution. Reading point #1 -3.4 degree celsius #2 -4.2 degree cescius Average: -3.8 degree celsius ΔTf: -13 degree celsius

Molality of Soultion:______________

Determine the Molar Mass of glycerin from experimental data. What value will you assign i? _1_

___________________

Finally, given that the formula for glycerin is C3H8O3, calculate % error in your determination?



Questions:

1. To what minimum temperature would a radiator be protected if equal volumes of ethylene glycol (density 1.11 g/mL) and water were mixed. Assume that ethylene glycol does not dissociate and has a formula C2H6O2. (assume you have a 2 L solution.



2. A 7.85 g sample of a compound with an emperical formula C1H1 is dissolved in 172g of benzene. The freezing point of the solution is 1.50 degree C below that of pure benzene. Assuming that the compound does not dissociate, what is the molar mass and molecular formula of this compound? The freezing point depression constant for benzene is 5.12 degree C/m.

This post was edited by alberster on May 2 2013 06:27pm

Will pay fgs for the solutions people!

V) Molality = ΔTf / Kf

Remember:
ΔTf is your freezing point lowering value: take the pure solvent freezing point minus the new freezing point of the solution after the solute (glycerin) is added.
This value will be positive, as you subtract a negative number. For example, if your solvent was water and glycerin was added, then you would take the original freezing point of water (0C) and subtract the new recorded freezing point (your average was -3.8C). Your ΔTf should come out to be 3.8C, in this case.

Kf is your Freezing-Point-Depression Constant. For water, this value is 1.858C/m.
The m, above, is molality.
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From molality, calculate the molar mass of glycerin. Determine the number of moles of glycerin that are 'dissolved' into the provided mass of the water.
For example (these values are not correct): 0.088 mol Glycerin / kg x (78.1 x 10^-6 kg water) = 6.9 x 10^-6 mol glycerin

Molar mass can be calculated by taking the mass of your glycerin (2.84 g) divided by the number of moles calculated above (once again, that value is not correct - use the data that you collected for this).
2.84 g glycerin / (6.9 x 10^-6 mol glycerin) = An incorrect and huge number based upon the mass of glycerin used for this example. Expect a value just below 100 g/mol, in your case.

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Calculating the % error: ( | Accepted - Experimental | / Accepted) x 100

or

Plug in your known value and your experimental value (the calculated molar mass of glycerin), accordingly, into the example below. It's the same basic formula that I provided above, but it's broken down into steps. It might be easier for you to understand.

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I'll work through #1 and #2 later, when I have some extra time, if no one else has attempted by then.


If you have any questions about the content that I covered above, please let me know. Try to apply the data that you collected from your experiment to the examples that I had provided. If you still have issues, I'll need some more information about your procedures in order to work out accurate answers for your problems. I could, in the information provided, make some assumptions, but I'd rather have you try to work it out first.

This post was edited by Fault on May 3 2013 03:33pm

1) Looks like you'll be solving for the point at which the solution will freeze.
Remember: ΔTf = Kf x m

Calculate the number of moles of solute, first:
number of moles of C2H6O2 = 1 L C2H6O2 (1.11 g / mL) (1000 mL / 1 L) (1 mol / 62.08 g) = 17.88 mol C2H6O2

Plug and solve for molality:
m = 17.88 mol / 2.11 kg = 8.47 molality

Plug and chug:
ΔTf = (1.86 C / m) (8.47 m) = 15.76 C

Now take the calculated temperature and subtract it from the freezing point of pure H2O (0C):
ΔTf = 0 C - 15.76 C = -15.76 C

The new solution will freeze at around -15.76 C.

2) Identify what information the problem gives you:
Mass of unknown solute = 7.85 g
Mass of solvent = 172 g C6H6
Kf of C6H6 = 5.12 C / m

F.P. of solution = 1.50C below pure C6H6
Therefore, ΔTf = 1.50C

Use the given information to solve for molality:
m = ΔTf / Kf = 1.50C / (5.12 C / m) = 0.293 m

Determine the number of moles of unknown solute in the solution:
0.293 mol / kg x (172 x 10^-3 kg C6H6) = 0.05 mol

Now calculate the molar mass of the unknown compound:
M.M. = 7.85 g / 0.05 mol = 155.78 amu

The Empirical formula was C1H1, and the mass of that formula is 13.02 amu.
Take the molar mass divided by the empirical formula mass: 155.78 amu / 13.02 amu = 12

The molecular formula, therefore, is: C12H12

It has been awhile since I've done these types of problems, but hopefully these are correct.
Please, once again, let me know if you have any questions, or if you would like more involved explanations.

This post was edited by Fault on May 3 2013 05:03pm