# Question #c3228

##### 1 Answer

You'd need **3 g** of urea.

#### Explanation:

The trick here is to realize that the *cryoscopic constant* for water is actually given to you **per 100 g** of water, instead of **per kilogram**.

#K_f = 18.6 "K"/("mol"/"100g") = "18.6 K mol"^(-1)"100g"#

The difference between these values lies in the fact that molality is defined as moles of solute, in your case urea, per **kilogram** of solvent, in your case water.

This means that you need to convert the value of the cryoscopic constant to help you determine the molality of the solution, which in turn will get you the number of moles of urea.

The cryoscopic constant will thus be

#"18.6 K mol"^(-1) 100color(red)(cancel(color(black)("g"))) * "1 kg"/(1000color(red)(cancel(color(black)("g")))) = "1.86 K mol"^(-1)"kg"#

The equation for *freezing-point depression* looks like this

#DeltaT_"f" = i * K_f * b# , where

**cryoscopic constant**;

**van't Hoff factor** - the number of ions per individual molecule of solute.

Urea is a nonelectrolyte, which means that the van't Hoff factor will be equal to **1**.

Use this equation so solve for

#b = (DeltaT_"f")/(i * K_f)#

#b = (0.186color(red)(cancel(color(black)("K"))))/(1 * 1.86color(red)(cancel(color(black)("K")))"mol"^(-1)"kg") = "0.1 mol kg"^(-1)#

You can assume the density of the solution to be equal to

This means that you have

#b = n_"urea"/m_"solvent" implies n_"urea" = b * m_"solvent"#

#n_"urea" = "0.1 mol" * color(red)(cancel(color(black)("kg"^(-1)))) * 500 * 10^(-3)color(red)(cancel(color(black)("kg")))#

#n_"urea" = "0.05 moles"#

Now use the compound's molar mass to determine how many grams would contain this many moles

#0.05color(red)(cancel(color(black)("moles"))) * "60.06 g"/(1color(red)(cancel(color(black)("mole")))) = "3.003 g"#

Rounded to one sig fig, the number of sig figs you gave for the volume of the solvent, the answer will be

#m_"urea" = color(green)("3 g")#