# Question #75a08

##### 1 Answer

#### Answer:

#### Explanation:

AN important part of this problem is the fact that you're dealing with a *molecular* compound. That means that it will dissolve in aqueous solution **without dissociating** into ions.

An important consequence of that is the fact that the *van't Hoff factor*, **produced in solution** by dissolving the solute, will be equal to

In simple words, one molecule dissolved, one molecule in solution.

Now, the equation that describes *freezing-point depression* looks like this

#color(blue)(DeltaT_f = i * K_f * b)" "# , where

*van't Hoff factor*, equal to

*cryoscopic constant* of the solvent;

The cryoscopic constant of water is equal to

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

Now, the freezing-point depression is defined as

#color(blue)(DeltaT_f = T_f^@ - T_f)" "# , where

*pure solvent*

In your case, the freezing-point depression will be

#DeltaT_f = 0^@"C"- (-0.104^@"C") = 0.104^@"C"#

So, all you have to do from this point on is to plug your values into this equation and solve for

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

#b = (0.104 color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.86 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = color(green)("0.0559 mol kg"^(-1))#