# Methanol, CH_3OH is used as an antifreeze. How many grams of methanol would you need per 1000.0 g of water for an aqueous solution to stay liquid at -20.0° C?

Dec 11, 2015

${m}_{C {H}_{3} O H} > \text{355 g}$

#### Explanation:

Your strategy here will be to use the equation for freezing-point depression to determine the molality of the target solution, then use methanol's molar mass to find the mass of the compound needed to make this solution.

The equation that describes freezing-point depression looks like this

$\textcolor{b l u e}{\Delta {T}_{f} = i \cdot {K}_{f} \cdot b} \text{ }$, where

$\Delta {T}_{f}$ - the freezing-point depression;
$i$ - the van't Hoff factor
${K}_{f}$ - the cryoscopic constant of the solvent;
$b$ - the molality of the solution.

The cryoscopic constant of water is equal to $1.86 {\text{^@"C kg mol}}^{- 1}$

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

Now, since you're dealing with a non-electrolyte, the van't Hoff factor, which tells you what the ratio between the number of moles of solute dissolved in water and the number of moles of particles produced, will be equal to $1$.

The freezing-point depression is defined as

$\textcolor{b l u e}{\Delta {T}_{f} = {T}_{f}^{\circ} - {T}_{f}} \text{ }$, where

${T}_{f}^{\circ}$ - the freezing point of the pure solvent
${T}_{f}$ - the freezing point of the solution

Pure water freezes at ${0}^{\circ} \text{C}$, which means that this solution's freezing-point depression will be equal to

$\Delta {T}_{f} = {0}^{\circ} \text{C" - (-20.0^@"C") = 20.0^@"C}$

Rearrange the equation for freezing-point depression and solve for $b$

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

b = (20.0 color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.86 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = "10.753 mol kg"^(-1)

As you know, molality is defined as moles of solute per kilograms of solvent. Use the given mass of water to find how many moles of methanol must be present in the solution

$b = {n}_{\text{solute"/m_"solvent" implies n_"solute" = b * m_"solvent}}$

n = "10.753 mol" color(red)(cancel(color(black)("kg"^(-1)))) * 1000.0 * 10^(-3)color(red)(cancel(color(black)("kg")))

$n = \text{10.753 moles CH"_3"OH}$

Finally, use methanol's molar mass to find how many grams would contain that many moles

10.753 color(red)(cancel(color(black)("moles CH"_3"OH"))) * "32.042 g"/(1color(red)(cancel(color(black)("mole CH"_3"OH")))) = "344.55 g"

Rounded to three sig figs, the number of sig figs you have for the freezing temperature of the solution

${m}_{C {H}_{3} O H} = \text{355 g}$

So, a solution that contains $\text{355 g}$ of methanol in $\text{1000.0 g}$ of water will freeze at $- {20}^{\circ} \text{C}$, which means that you'd need the mass of methnol to be bigger than this value

${m}_{C {H}_{3} O H} > \textcolor{g r e e n}{\text{355 g}}$