Exercise about an aqueous solution (containing a halogen)

[Ako9]

Homework Statement

"An aqueous solution, with 5% of CaX_2 (X is an halogen), freezes at t=-1.396°C. K_cr=1.860. Find X."

Relevant Equations

$$\Delta t = K_{cr}\cdot m \cdot i$$

I can't find my mistake, can you help me?

I suppose that in 1000g of solution there are 50g of solute and 950g of solvent.
Then I find the molality
$$m= \frac{50}{MM} \cdot \frac{1}{0.950}$$
So I can find the Molar Mass of the compound using the relation $$\Delta t=K_{cr}\cdot m \cdot i$$ where i is the van 't Hoff coefficient which is 3 in this case.
At the end I don't find an halogen (Rubidium is the closest) so I assume something is wrong but I can't see what.

 

[mjc123]

I think you've done the calculation correctly - at least, I agree with your answer. You get a halogen if you assume there are 50g solute to 1000g water. I'm never quite sure what people mean when they say "x% solution" with no further specification. Perhaps they think that this solution is sufficiently dilute that you can make this approximation, but that seems inconsistent with measuring T to 4 sig figs. (By the way, we usually use T for temperature, t for time.)

 

[Borek]

Last time I checked rubidium wasn't a halogen

 

[Mayhem]

The molality can be found exactly using the information given and the freezing point of water. You didn't use this number?

 

[Borek]

The molality can be found exactly using the information given and the freezing point of water. You didn't use this number?

They did, but it doesn't produce a reasonable answer. Turns out the 5% part is unreliable, try to do the calculations assuming 50/950 and 50/1000 and you will see where the problem is.

 

[Mayhem]

They did, but it doesn't produce a reasonable answer. Turns out the 5% part is unreliable, try to do the calculations assuming 50/950 and 50/1000 and you will see where the problem is.

I did the calculation and seem to be getting nonsense numbers. Concentration in % is honestly the devil.

 

[Borek]

% is OK as long as used correctly, it is just misused here.