Question #f5ef6
1 Answer
Here's what I got.
Explanation:
First of all, you're dealing with freezing-point depression, not freezing-point elevation.
Simply put, the freezing point of a solution will be lower than the freezing point of the pure solvent. The freezing-point depression tells you how low the freezing point of the solution will actually be compared with that of the pure solvent.
Now, the equation that allows you to calculate freezing-point depression looks like this
#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))#
Here
As you can see, the freezing-point depression cannot be negative because the van't Hoff factor, the cryoscopic constant, and the molality of the solution are all positive values.
In your case, aluminium nitrate,
#"Al"("NO"_ 3)_ (3(aq)) -> "Al"_ ((aq))^(3+) + 3"NO"_ (3(aq))^(-)#
Since one mole of solute produces
The cryoscopic constant of water is equal to
#K_f = 1.86^@"C kg mol"^(-1)#
The freezing-point depression will thus be
#DeltaT_f = 4 * 1.86^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.5 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#
#DeltaT_f = 3.72^@"C"#
This is the correct value for the freezing-point depression of a
Since freezing-point depression tells you by how many degrees the freezing point of the solution decreased compared to that of the pure solvent, which is
#T_"f sol" = 0^@"C" - 3.72^@"C" = -3.72^@"C"#
Now, the question asks you to find the freezing point of a solution in which ethanol is the solvent, not water.
The cryoscopic constant for ethanol is
#K_f = 1.99^@"C kg mol"^(-1)#
http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf
Assuming that you have complete dissociation, the freezing-point depression will be
#DeltaT_f = 4 * 1.99^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.5 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#
#DeltaT_f = 3.98^@"C"#
This means that the freezing point of the solution will be
#T_"f sol" = -18^@"C" - 3.98^@"C" = -21.98^@"C"#
So remember, the freezing-point depression is always positive and it must be subtracted from the freezing point of the pure solvent in order to get the freezing point of the solution.