When 9.12 g of HCl was dissolved in 190 g of water, the freezing point of the solution was -4.65oC. What is the value of the van't Hoff factor for HCl?
Here is what I did-
I know that I use: deltaT= Kf x m x i
rearranged to solve for i: deltaT/Kf x m
solving for molality:
mol HCl=0.250
kg solvent= .190
m=0.250/.190= 1.32
i= -4.65oC/1.86 x 1.32 = -1.89
I am not sure if the value for i can be negative or if I did the problem correctly. Please help me! Thanks!
Although for this question, you will have to determine the Van't Hoff factor using the formula, there is a neat of verifying your answer.
Let us denote the Van't Hoff factor by i.
The i value for a molecule like say glucose (does not dissociate when dissolved in a solvent like water)is generally considered as 1.
The i value for a strong base or acid or salt is obtained by adding the number of anions and cations.
For eg: MgCl2 has Mg+2 and 2 Cl- thereby a total of 3 ions.
But for a weak acid or a weak base i is considered to be 1.
Now for HCl, since it is a strong acid and is dissociated as H+ and Cl-, we can expect it to have i value as 2 and the answer that you obtained as 1.89 is fairly close to 2.0
Hope that additional explanation helps.
the Van't Hoff factor is always positive. Kf for water is -1.86 Kelvin/molal. Your works is otherwise correct, you're just missing the negative sign on Kf. Hope this helps.
Even though the freezing point is depressed (negative), i is usually positive, or its absolute value. In some literature it has been negative. From the equation you see that a small amount of HCl causes a significant temperature change. Good property for an anti-freeze. However HCl would rapidly destroy the inside of an engine so not a good choice.
As long as you show your work, if you do include a negative sign there should be no penalty. It shows you were thinking that the freezing point dropped. Be aware that the boiling point is elevated.
NaCl which is added to water to cook live crabs will elevate the boiling point. Recently we were cooking crabs and my son added "too much" salt. We waited a very long time for that water to boil. Not part of your problem, but when you can apply it to everyday life it becomes more relevant.
- George C. 12/2/2012
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Comments
(delta)f{initial freezing-final freezing point as depression of freezing point occurs } =iKfM —> i = deltaf/KfM delta f = 0 - (-4.65) = +4.65 so i is +... actually we always take temperature in kelvin and freezing point of water is 272.15k and new freezing point is 267.50k.
- Nikhil R. from Mankato, MN 2/5/2013