Salt Water Freezing Point Depression
Most recent answer: 01/27/2012
Q:
hi
were are doing a experiment and I was wondering what would happen if salt in ice affects how easily it would melt at room temp
- nichola (age 14)
london
- nichola (age 14)
london
A:
Salt, or anything else dissolved in water, will lower the freezing temperature of the water it is dissolved in. Really any solvent's freezing point will lower when it has something dissolved in it. This makes it harder to freeze warm salt water, and easier to melt already frozen salted ice.
Here's an equation for it: ΔT=Kf*b*i
here ΔT is the change in freezing point in Kelvins (Celcius will work too since they have the same scale).
Kf is the cryoscopic constant of the solvent, different for every material. For water Kf=1.853 K*kg/mol.
b is the molality of the solution, which is the amount of solute expressed in mols per kilogram of solvent.
i is called the van't Hoff factor. This is the number of ions in solution for each dissolved molecule. For salt, NaCl this number is 2
So if we dissolve 1.5 mol (87.75g) of table salt in 1 Kg (1L) of water our equation is:
ΔT=1.853 * 1.5 * 2 = 5.5 K
Since table salt forms an ionic solution, if you start to add too much this equation starts to differ from the actual phenomenon a bit, but as long as its a relatively small amount this should fit pretty well.
Thanks for the question,
Mike Boehme
Here's an equation for it: ΔT=Kf*b*i
here ΔT is the change in freezing point in Kelvins (Celcius will work too since they have the same scale).
Kf is the cryoscopic constant of the solvent, different for every material. For water Kf=1.853 K*kg/mol.
b is the molality of the solution, which is the amount of solute expressed in mols per kilogram of solvent.
i is called the van't Hoff factor. This is the number of ions in solution for each dissolved molecule. For salt, NaCl this number is 2
So if we dissolve 1.5 mol (87.75g) of table salt in 1 Kg (1L) of water our equation is:
ΔT=1.853 * 1.5 * 2 = 5.5 K
Since table salt forms an ionic solution, if you start to add too much this equation starts to differ from the actual phenomenon a bit, but as long as its a relatively small amount this should fit pretty well.
Thanks for the question,
Mike Boehme
(published on 01/27/2012)