[DannyBoi]

Hi guys,

I'm working on a freezing point depression prelab where we determine the freezing point of a pure solvent then compare  with its freezing point in a solution with an unknown solute. This ΔT will be used in the ΔT_fp = K_fpm_solute to obtain the moles of the solute. Knowing the mass and moles of solute will give us its molar mass, which will allow us to obtain the molecular formula of the unknown.

The 2nd question asks:

Derive a formula for the molar mass of the solute in terms of the freezing point
depression, ΔT, the freezing-point depression constant, K_f, and the masses of solvent,
m_solvent, and solute, m_solute, used in an experiment like this one. Show your work, and
include all units in your final formula.

This stumped me a bit because I'm not sure what it's asking exactly. After playing around with units and the above equation I came up with:

(K_fpm_solvent(kg))/ΔT(m_solute(g))= molar mass solute

I did this honestly just trying to make the units fit, and they do as far as I can see. But is this what the question was asking?

[DannyBoi]

Sorry I meant:

(K_fpm_solvent(kg)m_solute(g))/ΔT= molar mass solute

[opti384]

Try again. Start from ΔT = K_fpm(molality)

[Borek]

What is molality definition?

[DannyBoi]

So what I'm trying to do is rewrite the ΔT = K_fpm_solute equation so that one of the sides will be g/mol_solute? Can I really do that without adding anything to the equation?

Molality as far as I know is mol_solute/kg_solvent. And K_fp is in units of (^oC/(mol/kg)). So then I would guess that the units work out like this:

ΔT(^oC) = K_fp(^oC)(mol)/(mol/kg)(kg) which I can try to rewrite...

ΔT(^oC)/(K_fp(^oC)/(mol/kg)) = (mol)/(kg) Do you times both sides by 1kg/1000g to get it in molar mass units?

But isn't the kg in these equations referring to the mass of the solvent and not the solute? How can you write such an equation in terms of the mass of solute when it's not in the original ΔT = K_fpm_solute

I'm doing all this blind but to be honest I'm not really aware of the correct process or what it means, it's a bit frustrating because I do want to understand. Sorry for the long winded posts haha.

[Borek]

Ignore this post, it was a correction but I just realized you can modify your original posts.

Only for a limited period of time.

Why don't you start with molality definition, put it into your ΔT equation, then solve for molar mass?

[DannyBoi]

Hey Borek!

Molality would be mol_solute/kg_solvent.

So didn't I solve the equation for molar mass in my last post?

ΔT(^oC) = K_fp(^oC)(mol)/(mol/kg)(kg)

ΔT(^oC)/((K_fp(^oC)/(mol/kg)) = (mol)/(kg)

and then you times both sides by 1kg/1000g to get an equation that equals something with units mol/g which are the units for molar mass. Guess that's probly not what you mean?

[Borek]

Sorry, but the way you are doing it is hard to follow. Don't use units, use symbols, guessing what you mean by (mol)/(mol/kg)(kg) is a waste of time.

[DannyBoi]

Ok, I saw that the K_fp in my chem text was in units of ^oC/m so I put the m in the denominator and turned it into mol/kg. I really am trying to understand, but I guess I will need to wait to see what it was asking for when I hand it in and get it back.

[Borek]

You should try to solve using some easy to follow symbols. Units are not for SOLVING, units are for checking the result correctness. Sometimes it is possible to play with units to find a possible suggestions as to how the answer may look like, but this is not a general method of solving simple questions.

Assume molar_mass is molar mass of the solute.

Then



and molality



Now we can write



and solve for molar mass - and thats it.

Geez, this LaTeX looks terrible

[DannyBoi]

Wow ok, it definitely is much more intuitive and clear when you use symbols. I should have known to rewrite mol_solute in such a manner! These questions usually end up so much simpler than how my mind confuses them! HAHA

Thanks for the insight!