# Urea, #("NH"_2)_2"CO"#, is dissolved in #"100 g"# of water. The solution freezes at #-0.085^@"C"#. How many grams of urea were dissolved to make this solution?

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(Given #K_f# of water = 1.858 #K_b# = 0.512)

(Given

##### 1 Answer

#### Answer:

#### Explanation:

Your tool of choice here will be the equation that allows you to calculate the **freezing point depression** of the solution

#color(blue)(ul(color(black)(DeltaT_"f" = i * K_f * b)))#

Here

#T_"f"# is thefreezing point depression#i# is thevan't Hoff factor#b# is themolalityof the solution#K_f# is thecryoscopic constantof water, equal to# 1.858^@"C kg mol"^(-1)#

Now, the **freezing point depression** tells you the difference between the freezing point of the pure solvent and the freezing point of the solution.

In other words, the freezing point depression tells you by how many degrees the freezing point of the solution **decreases** compared with the freezing point of the pure solvent.

You know that your solution freezes at **freezing point depression** will be

#DeltaT_"f" = 0^@"C" - (-0.085^@"C")#

#DeltaT_"f" = 0.085^@"C"# This essentially means that the freezing point of the solution is

#0.085^@"C"# lowerthan the normal freezing point of the pure solvent.

So, rearrange the equation to find the **molality** of the solution.

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

Since urea is a **nonelectrolyte**, which means that it does not dissociate in aqueous solution, its van't Hoff factor will be equal to

This means that you have

#b = (0.085 color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.858 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = "0.04575 mol kg"^(-1)#

This tells you that this solution contains **moles** of urea for every

#100 color(red)(cancel(color(black)("g water"))) * "0.04575 moles urea"/(10^3color(red)(cancel(color(black)("g water")))) = "0.004575 moles urea"#

Finally, to convert the number of grams to *moles*, use the **molar mass** of urea

#0.004575 color(red)(cancel(color(black)("moles urea"))) * "60.06 g"/(1color(red)(cancel(color(black)("mole urea")))) = color(darkgreen)(ul(color(black)("0.27 g")))#

I'll leave the answer rounded to two **sig figs**, but don't forget that you have one significant figure for the mass of water.