# CHEMISTRY

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Assuming equal concentrations, rank these aqueous solutions by their freezing point: LiSO4, Li3PO4, SnCl4, NH4Br.
(from highest freezing point to lowest freezing poinT)

Thanks.

• CHEMISTRY - ,

delta T = i*Kf*m is the equation you will need. You don't have any m listed but you know all are the same. Kf is the same for all. So the only difference is i may be different. i is the van't Hoff factor and it's the number of particles when a salt dissociates. First, look at your notes or post again, LiSO4 should be Li2SO4.
Li2SO4 is 3 for it dissociates as Li2SO4 ==> 2Li^+ + SO4 (3 particles).

• CHEMISTRY - ,

yeah i meant Li2SO4

• CHEMISTRY - ,

So how would i determine it...based on the van't Hoff factor?

• CHEMISTRY - ,

If m is constant and Kf is constant, then delta T depends ONLY upon i, right? So i goes up, delta T goes (up, down). i goes down, delta T goes (up,down).
So you look at all of the salts, determine i for each and delta T for each. Then you will know which salt provides the lowest f.p.

• CHEMISTRY - ,

So the order from highest to lowest is :Li3PO4, NH4Br, SnCl4,Li2SO4 .... please confirm the right answer.

• CHEMISTRY - ,

No.
Li2SO4; i = 3 Li2SO4 ==> 2LI^+ + SO4^2-
Li3PO4; i = 4 Li3PO4 ==L> 3Li^+ + PO4^3-
NH4B4; i = 2 NH4Br ==> NH4^+ + Br^-
SnCl4; i = 5 SnCl4 ==>: Sn^4+ + 4Cl^-

Since delta T = i*Kf*m and the only variable is i, then the lowest i will have the lowest delta T (and the highest freezing point) while the highest i will have the largest delta T and the lowest freezing point. Rank them 2,3,4,5,
I would rank them from highest f.p.(lowest delta T) to lowest f.p.(highest delta T) as NH4Br, Li2SO4, Li3PO4, SnCl4.
Remember that freezing point pure H2O is 0 C. If delta T = 1 then freezing point of the new soln will be 0 - 1 = -1. If delta T = 2 the freezing point of the new soln will be 0 - 2 = -2. If delta T = 5 the freezing point of the new soln will be 0 - 5 = -5
You can see that the lowest f.p. (-5) is the largest delta T.

• CHEMISTRY - ,

Oh gotcha! Thanks Dr Bob!!! :D I really appreciate it!