# 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?

## (Given ${K}_{f}$ of water = 1.858 ${K}_{b}$ = 0.512)

Jan 15, 2018

$\text{0.27 g}$

#### Explanation:

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

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\Delta {T}_{\text{f}} = i \cdot {K}_{f} \cdot b}}}$

Here

• ${T}_{\text{f}}$ is the freezing point depression
• $i$ is the van't Hoff factor
• $b$ is the molality of the solution
• ${K}_{f}$ is the cryoscopic constant of water, equal to${1.858}^{\circ} {\text{C kg mol}}^{- 1}$

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 $- {0.085}^{\circ} \text{C}$. Since pure water has a normal freezing point of ${0}^{\circ} \text{C}$, you can say that the freezing point depression will be

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

$\Delta {T}_{\text{f" = 0.085^@"C}}$

This essentially means that the freezing point of the solution is ${0.085}^{\circ} \text{C}$ lower than the normal freezing point of the pure solvent.

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

$b = \frac{\Delta {T}_{\text{f}}}{i \cdot {K}_{f}}$

Since urea is a nonelectrolyte, which means that it does not dissociate in aqueous solution, its van't Hoff factor will be equal to $1$.

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 $0.04575$ moles of urea for every $\text{1 kg" = 10^3 quad "g}$ of solvent. Since your sample contains $\text{100 g}$ of water, you can say that it will also contain

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

$0.004575 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles urea"))) * "60.06 g"/(1color(red)(cancel(color(black)("mole urea")))) = color(darkgreen)(ul(color(black)("0.27 g}}}}$

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.