## anonymous 4 years ago 14 GRAMS of an electrolyte was dissolved in 41 grams of water, the resulting solution was found to have a molarity of 0.397 mol / L. The freezing point of the solution was determined to be -9.18°C. The freezing point depression constant for water is -1.86°C and pure water may be assumed to freeze at 0°C. If the molecular weight of the electrolyte is 130.3 g/mol and it takes 45 minutes to perform the experiment, what is the freezing point of the solution ? (°C) Can someone explain how to do this?

1. Rogue

Use the freezing point depression formula to do this. Delta T is your temperature difference, which is given to be -9.18°C. i is the van't Hoff factor, which is the # of ions the solute dissociates into. m is your molality. $\Delta T = i k_{f} m$$M = \frac {mol solute}{liters solution} = frac {n}{V}$$m = \frac {moles solute}{kilograms solvent}$

2. Rogue

$M = \frac {n}{V}$

3. Rogue

The thing that I'm not sure on is what value to use for the van't Hoff factor. The solute is an electrolyte, but the question does not specify its strength...

4. Xishem

You should be able to calculate the van't Hoff factor using the freezing point depression equation, no? You have the freezing point depression amount, the freezing-point-depression constant, and the molal concentration.

5. Xishem

Wait. The question is answered in the question, is it not? "The freezing point of the solution was determined to be -9.18°C." "What is the freezing point of the solution?"

6. Rogue

Yeah, that's what I'm wondering about now...

7. Xishem

It doesn't mention another solution, so I feel like this question is either a trick, or it's miswritten.

8. Rogue

Nkenna is that the whole question?

9. Preetha

I am also surprised at the question. The FP is given. There is too much information. Sounds like it is supposed to test you on whether you can sift through all the material.

10. anonymous

"It takes 45 minute to perform the experiment" hahahaha....

11. anonymous

Yes it is the whole question and maybe because I haven't learned anything that you guys have said is what is confusing me

12. anonymous

I think the answer is -9.18 as well but I wasn't sure since they gave all this other information so I'm not sure

13. Xishem

It's very common in any science to test the student's knowledge of sifting through information. It seems, to the best that we know, that that is what this problem is doing.