# The freezing point of an aqueous solution that contains a nonelectrolyte is #-4.0^@"C"#. What is the molal concentration of the solution? #Kf = 1.86^@"C/m"#

##### 1 Answer

#### Explanation:

The first thing to do here is use the freezing point of the **solution** to calculate the *freezing-point depression*,

As you know, the **freezing-point depression** is a measure of how **low** the freezing point of a solution is compared with the freezing point of the *pure solvent*.

You're dealing with an * aqueous solution*, so right from the start you know that the solvent is water. Pure water freezes at

Your solution freezes at **decrease** in the freezing temperature of the solution compared with that of pure water

#color(purple)(|bar(ul(color(white)(a/a)color(black)(T_"f sol" = T_"f pure solvent" - DeltaT_f)color(white)(a/a)|)))#

#-4.0^@"C" = - DeltaT_f implies DeltaT_f = 4.0^@"C"#

Now, the freezing-point depression is calculated sing the equation

#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;

Your solute is a **non-electrolyte**, which means that its molecules * do not* dissociate in aqueous solution. This implies that the van't Hoff factor will be equal to

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 find

#b = (4.0 color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.86 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = color(green)(|bar(ul(color(white)(a/a)color(black)("2.2 mol kg"^(-1))color(white)(a/a)|)))#

The answer is rounded to two **sig figs**, the number of sig figs you have for the freezing point of the solution.