# 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

*van't Hoff factor*

*cryoscopic constant* of the solvent;

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, **soluble** in aqueous solution, which means that it dissociates completely to form aluminium cations,

#"Al"("NO"_ 3)_ (3(aq)) -> "Al"_ ((aq))^(3+) + 3"NO"_ (3(aq))^(-)#

Since one mole of solute produces **moles** of particles of solute in solution, the van't Hoff factor is equal to

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 **water is the solvent**.

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 **the solution** will be

#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 **lower** than the freezing point of the pure solvent, which for ethanol is

#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**.