calculating the van't Hoff factor

Question

When 9.12 g of HCl was dissolved in 190 g of water, the freezing point of the solution was -4.65^oC.  What is the value of the van't Hoff factor for HCl?

Here is what I did-

I know that I use: deltaT= K_f x m x i

rearranged to solve for i: deltaT/K_f x m

 

solving for molality:

mol HCl=0.250

kg solvent= .190

m=0.250/.190= 1.32

 

i= -4.65^oC/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!

Answer 1

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.

Answer 2

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.

 

Answer 3

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.