# What is the molality of #NaCl# if the freezing point of a #NaCl# solution is -5.58 C°?

##### 1 Answer

#### Explanation:

A solution's **molality** will tell you how many *moles of solute* you get **per kilogram of solvent**.

In this case, the only information you have is that a given sodium chloride solution has a *freezing point* of

As you know, the freezing point of a solution is a **colligative property**, meaning that is depends exclusively on the **concentration of solute particles**.

Since sodium chloride is an **electrolyte**, it will dissociate completely in aqueous solution to form sodium cations and chloride anions

#"NaCl"_text((aq]) -> "Na"_text((aq])^(+) + "Cl"_text((aq])^(-)#

Notice that **one mole** of sodium chloride produces **one mole** of sodium cations and **one mole** of chloride anions in solution, i.e. **two moles of particles**.

This mans that sodium chloride has a **van't Hoff factor**, *per mole* of solute dissolved in aqueous solution.

Now, you can *compare* the freezing point of the solution with that of *pure water*, which is equal to **freezing-point depression** of the solution,

#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = T_f^@ - T_"f sol"color(white)(a/a)|)))" "# , where

**pure solvent**

In your case, the freezing-point depression will be equal to

#DeltaT_f = 0^@"C" - (-5.58^@"C") = 5.58^@"C"#

The freezing-point depression is also related to the **molality** of the solution

#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))" "# , where

*van't Hoff factor*

*cryoscopic constant* of the solvent;

The cryoscopic constant of water is equal to

http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf

Rearrange the above equation to solve for

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

Plug in your values to get

#b = (5.58color(red)(cancel(color(black)(""^@"C"))))/(color(red)(2) * 1.86color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = color(green)(|bar(ul(color(white)(a/a)"1.50 mol kg"^(-1)color(white)(a/a)|)))#

The answer is rounded to three **sig figs**.