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calculating the van't Hoff factor

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!

(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.

2/5/2013

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: MgClhas 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

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.