[cseil]

So I've got this problem:

The freezing point of water in equilibrium with ethyl ether is -3.853°C.
The freezing point of water in equilibrium with a solution of 10.5g of the hydrocarbon N in 100g of ether is 0.232°C higher that the other case (so, -3.621°C).

What's the mass percentage of ether in a solution near the freezing point?
What's the molecular weight of N?

N is not soluble in water and the solubility of the ether doesn't change with the temperature.
Kcr for water is 1.86.

The first part is easy:
I calculate m from [tex]\Delta T = K_{cr}*m[/tex]
and I get m=2.06 mol/kg

2.06 moles of ethyl ether are 153.18g and the mass percentage is 13.28%.

I cannot understand the second part.
If I apply the same equation with a different ΔT I get m=1.947mol/kg.

But I know that the solution is made by 100g of ether, so:

[tex]\frac{1.35mol}{x kg} = 1.947[/tex]

x = 693g (water).

The solution is 693g water, 100g ether, 10.5g N.
I can get the mass percentage but I don't know how to go on with the problem.

Can you help me?

Thank you

[mjc123]

What does "water in equilibrium with ether" (or a solution of N in ether) mean? It means the solutions are saturated - you have water (saturated with ether) in contact with ether (saturated with water). In the second case, you have two solutions - water (saturated with ether) and a solution of N in ether (saturated with water). How is the chemical potential of ether affected by dissolving N?

[cseil]

Considering that the chemical potential is

[tex]\mu = \mu^* + RTln(x_i)[/tex]

if I dissolve N in ether, the μ will be lower and the difference is in the x_i, right?

[mjc123]

Yes

[cseil]

Ok here's my reasoning now:

I calculated the mass of every component of the solution in the first post:

693g of water, 100g of ethyl ether, 10.5g of N
38.46moles water, 1.349mol ether, 10.5g/Mx of N

I have to find Mx.

I converted the molality to the molar fraction with the approximated equation:
[tex]m=\frac{1000}{M_0} x[/tex]

and I have:
x1 (first case)=0.0373
x2 (second case)=0.0351

The molar fraction of N is Δx = 2.2x10^-3.

[tex]x_N = \frac{10.5}{M_x (38.46 + 1.349 + \frac{10.5}{M_x})}[/tex]

and I find Mx=119.6 g/mol.

The book says 121.5g.

EDIT Now I remember about another topic in which you said that the molar fraction doesn't vary linearly with molality. You said I had to write x/(1-x) and not x. But I'm thinking about that and I cannot understand why.

[mjc123]

I calculated the mass of every component of the solution in the first post

There isn't "a solution". There are two immiscible solutions in equilibrium with each other. That was my point.
As for the equation, if you have a solution of molality m mol/kg
moles solute (1) = m
moles solvent (2) = 1000/M₂
x₁/x₂ = x₁/(1-x₁) = moles1/moles2 = mM₂/1000
If x₁ is very small (like ca. 0.03) the difference between x and x/(1-x) will not be great, but in your earlier post x was about 0.26, and the difference matters.

[cseil]

Eheh, sorry I know they are two solutions in equilibrium, I called it the wrong way 
Anyway, got it! Thanks!