# Find the freezing point (in ""^@ "C") for a solution consisting of "60.0 g" water and "40.0 g" ethylene glycol?

## Find the freezing point (in $\text{^@ "C}$) for a solution consisting of $\text{60.0 g}$ water and $\text{40.0 g}$ ethylene glycol? The ${K}_{f}$ of water is ${1.86}^{\circ} \text{C"cdot"kg/mol}$ the ${K}_{b}$ of water is ${0.512}^{\circ} \text{C"cdot"kg/mol}$, and the molar masses of water and ethylene glycol are $\text{18.015 g/mol}$ and $\text{62.0 g/mol}$, respectively. a) ${20.1}^{\circ} \text{C}$ b) $- {20.1}^{\circ} \text{C}$ c) ${5.50}^{\circ} \text{C}$ d) ${28.42}^{\circ} \text{C}$

##### 2 Answers
Sep 10, 2016

B -20.1 degrees C

#### Explanation:

The freezing point depression for water is -1.85 degrees per molal solution.

A molal solution is l mole of particles per kg of solution.

As ethylene glycol is a covalent molecule it does not ionize in water. Because of this 1 mole of ethylene glycol produces 1 mole of particles.

To find the number of moles of ethylene glycol divide the mass of ethylene glycol by the molar mass.

$\frac{40}{62}$ = ,645 moles.

There are 1000 grams of water in a Kg. To find the number of Kg of solution define the 60 grams grams of water by 1000 grams.

$60 \frac{g}{1000} g$ = .06 Kg.

The molal concentration is the moles divided by the Kg.

,645 moles / .06 Kg = 10.8 m ( molal concentration)

The freezing point depression is -1.85 ${C}^{o}$ per 1 -molal.
To find the depression multiply -1.85 x 10.8 molal.

$- 1.85 \times 10.8$ = - 19.88

The normal freezing point is ${0}^{o} C$ so the new freezing point is
0 + - 19.88 = - 19.88

The closest answer is B.

Sep 10, 2016

I got $\text{b}$.

a) is incorrect because freezing point depression is negative.
c) is incorrect because it results from using ${K}_{b}$ instead of ${K}_{f}$, and yet the change in boiling point is added to the freezing point...
d) is incorrect because it results from having calculated the molality for water instead of for ethylene glycol, and then used ${K}_{b}$ instead of ${K}_{f}$, and then still added the change in boiling point to the freezing point.

The ${K}_{f}$ of water is $\text{1.86"^@ "C/m}$, and recall the freezing point depression equation:

$\boldsymbol{\Delta {T}_{f} = - i {K}_{f} m}$

where:

• $i$ is the van't Hoff factor, which is the number of ions in a fully ionic compound. This was not given so we have to work from a different equation or estimate $i$.
• ${K}_{f}$ is the freezing point depression constant.
• $m$ is the molality of the solution, i.e. $\text{mols solute"/"kg solvent}$.
• $\Delta {T}_{f} = {T}_{f} - {T}_{f}^{\text{*}}$ is the change in freezing point, where ${T}_{f}^{\text{*}}$ is the freezing point of the pure solvent and ${T}_{f}$ is the new freezing point.

Ethylene glycol is not fully ionic, but it is polar, so it is miscible in water. However, its $\text{pKa}$ is significantly higher than that of water ($25$ vs. $15.7$), so we can say it does not dissociate significantly in water. Therefore, $i \approx 1$. It's probably a little higher than $1$ for a real solution, however.

The molality of the solution is based on which substance is the solvent, i.e. which one there is more of.

$\text{60.0 g"/"18.015 g/mol" = "3.331 mol water}$
$\text{40.0 g"/"62.0 g/mol" = "0.6452 mol EG}$

Therefore, water is the solvent.

color(green)(m_"soln") = "mols EG"/"kg water"

= "0.6452 mols EG"/(60.0 cancel"g water" xx "1 kg water"/(1000 cancel"g")

$=$ $\textcolor{g r e e n}{\text{10.75 m solution}}$

Therefore, the change in freezing point is:

$\Delta {T}_{f} \setminus = {T}_{f} - {T}_{f}^{\text{*" ~~ (1)("1.86"^@ "C"cdotcancel"kg/mol")(10.75 cancel"mol/kg") ~~ -20.0^@ "C}}$

Since the freezing point must decrease, the final freezing point is:

$\textcolor{b l u e}{{T}_{f}} = - {20.0}^{\circ} \text{C" + T_f^"*}$

$= \textcolor{b l u e}{- {20.0}^{\circ} \text{C}}$

The apparent answer is therefore $- {20.1}^{\circ} \text{C}$, which is $\text{B}$. None of the other answers are right because they are positive and not negative.

Based on this answer, can you see that $i \approx 1.005$? It was a good estimate to say that $i \approx 1$, but it's not exactly $1$.