Question

We actually talked about this before just to remind you im posting the previous post below:

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And i have a different question regarding a post someone else put up yesterday:
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Hey guys I did a lab and I have to answer the questions relating to the lab. I did question 1 & 2, can someone check if they are correct, and question 3 I don't know how to do it.

1. Calculation to Determine the molecular weight of unknown substance

Mass of unknown used: 2.0 g
Mass of water used: 50.0 g (0.05kg)
Kf = 1.86°C kg/mole
Experimentally determined freezing point of unknown: -1°C

Solution = delta Tf/-Kf
= -1°C/-1.86°C kg/mole
= 0.538 m (molality)

Molar mass = 2g / 0.05 kg * 0.538 m
= 74.4 g/mol


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These are my values:

Mass of unknown used:1.9397 g
Mass of water used: 50.0 g (0.05kg)
Kf = 1.86°C kg/mole
Experimentally determined freezing point of unknown: -.10°C

I did this lab as well, So i was wondering when it says:

Solution = delta Tf/-Kf

Wouldn't delta be final temp - initial temp:

So my initial temp was 23.00C and final -.10:

So : is delta T = -23.1


Solution = delta Tf/-Kf
= -23.1°C/-1.86°C kg/mole
= 12.4 m (molality)

Molar mass = 1.9397g / 0.05 kg * 12.4 m
= 481.04 g/mol
Which is totally wrong so now i m confused

Responses

* Chemistry------DrBobb.. - DrBob222, Saturday, February 7, 2009 at 12:21pm

With regard to question #2.
Yes, this was a previous post which I didn't answer but Bob Pursley did. He suggested that the student didn't have good precision in the measurements. In answer to your question, yes, delta T is 23.1 if you started with 23.00 and ended up with -0.1 (but it's +23.1 if that makes a difference because 23.00-(-0.1) = 23.00 + 0.1 = 23.1 degrees T. Next, if I'm able to separate what the other post was from your data, then I don't get your answer of 481.04 but my answer is ridiculous.
I agree with 12.4 m
Then m = mols/kg = 12.4 = mols/0.05
Solve for mols = 12.4 x 0.05 = 0.62
mols = g/molar mass = 0.62 = 1.9397/molar mass
Solve for molar mass = 1.9397/0.62 = 3 something and you know that can't be right. I don't remember the details of the problem but one possible problem could be that ionic compounds have an i in the equation (for number of particles). The second problem is simply that it's a lab procedure and we have no way of knowing how well the numbers have been measured. One comment I have, however, is that 12.4 m seems to be a high value for a 2 g sample in 50 mL water. But then I'm not familiar with the experiment and how it was conducted.

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NOW my question is i was looking around on website and found this:

This experiment was done using Napthalane instaead of water what i used in my experimnet:



What is the molecular weight of urea if the freezing point of 15g urea in 100g naphthalene is 63.3oC? The freezing point of the pure naphthalene is 80.6oC.

DT =Kfpm

m = DT/Kfp

Kfp for naphthalene is 6.9oC/m

DT= 80.6oC-63.3oC = 17.3oC

m=17.3oC/(6.9oC/m)

= 2.5 m = 2.5 mol urea/kilogram naphthalene

Remember that molality is the number of moles of solute per 1000 g (1 kilogram) of solvent. In this solution, the concentration is 15g urea in 100 g of naphthalene or 150 g urea in 1000 g of naphthalene.

Thus:

150g=2.5 mol

1mol=60 g

The molecular weight of urea is 60 g/mol.


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So they subtract the freezing point of the unknown from the freezing point of Napthalane:

SO i was to do it that way, the freezing point of water i determined when i did the experiment was 1.00 C.

SO :

Delta T: 1.00-(-.10)= 1.1

1.1= 1.86*m
0.59=m

Molar mass = 1.9397g / 0.05 kg * 0.59 m
= 65.75 g/mol

Which seems about right, because the other way i get molar mass= 3.12g:

So, would this be correct or should i stick to 3.12 because i had to hand in my calculations, so double checking.


Thank You In advance

Answer 1

Delta T is the difference between the normal freezing point and the freezing point of the solution. I wondered how you obtained a freezing point of 23 BUT I figured you had somehow managed it so 23 + 0.1 was the difference. If the freezing point of the solution before the solute was added is 0 degrees C and the freezing point with the solute added is -1.00, then delta T is 1. I didn't go through your calculations; I'm just commenting on your question.

Answer 2

the 23 i was using my initial temp of the solution..thats what i used