# Question #4859e

##### 1 Answer

#### Answer:

#### Explanation:

Even without doing any calculations, you could say that the answer should be

#"Na"_2"SO"_text(4(s]) -> 2"Na"_text((aq])^(+) + "SO"_text(4(aq])^(2-)#

This means that, in *theory*, the **van't Hoff factor**, which gives you the number of particles produced when **one molecule or formula unit**, depending if you have a covalent or an ionic solute, of the solute dissolves in the solvent, will be equal to

So, the equation that gives you the *freeezing-point depression* is

#DeltaT_f = i * K_f * b " "# , where

*van't Hoff factor*;

**cryoscopic constant** of the solvent;

The *freezing-point depression* is defined as

#DeltaT_f = T_"f"^0 - T_"f sol"#

In your case, you know that the freezing point of the solution is equal to

#DeltaT_f = 0^@"C" - (-4.218""^@"C") = 4.218""^@"C"#

You know that the *Cryoscopic constant* of water is equal to

#K_f = 1.853""^@"C kg mol"""^(-1)#

Rearrange the equation and solve for

#i = (DeltaT_f)/(K_f * b)#

#i = (4.218color(red)(cancel(color(black)(""^@"C"))))/(1.853color(red)(cancel(color(black)(""^@"C"))) color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-)))) * 0.8402color(red)(cancel(color(black)("mol")))/color(red)(cancel(color(black)("kg")))) = color(green)(2.709)#

The van't Hoff factor turned out to be *smaller* than the predicted value.

Since the solute does not produce as many particles per formula unit as it would produce **if completely dissolved**, you can conclude that not all the ions are completely separated from each other.

In other words, some sodium cations will actually bind to the sulfate anions and exist as *for every* sodium sulfate formula unit, you get something like

#"Na"_2"SO"_text(4(s]) -> 2"Na"_text((aq])^(+) + "SO"_text(4(aq])^(2-)#

and

#"Na"_2"SO"_text(4(s]) -> "Na"_text((aq])^(+) + "Na"^(+)"SO"_text(4(aq])^(2-)#