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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.

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