# What is the freezing point of an a nonionizing antifreeze solution containing 38.8g ethylene glycol, C_2H_6O_2, and 409 g of water?

Nov 20, 2015

${T}_{\text{f sol" = -2.84^@"C}}$

#### Explanation:

Your tool of choice for this problem will be the equation for freezing-point depression, which looks like this

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

$\Delta {T}_{f}$ - the freezing-point depression;
$i$ - the van't Hoff factor, equal to $1$ for non-electrolytes;
${K}_{f}$ - 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

In order to be able to determine the freezing point of this solution, you will first need to find its molality, which as you know is defined as the number of moles of solute divided by the mass of the solvent - expressed in kilograms!

Use ethylene glycol's molar mass to determine how many moles you have in that sample

38.8color(red)(cancel(color(black)("g"))) * ("1 mole C"_2"H"_6"O"_2)/(62.07color(red)(cancel(color(black)("g")))) = "0.625 moles C"_2"H"_6"O"_2

This means that the molality of the solution will be

$\textcolor{b l u e}{b = \frac{n}{m}}$

$b = \text{0.625 moles"/(409 * 10^(-3)"kg") = "1.528 molal}$

This means that the freezing-point depression for this solution will be

$\Delta {T}_{\text{f" = 1 * 1.86""^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 1.528color(red)(cancel(color(black)("moles"))) color(red)(cancel(color(black)("kg"^(-1)))) = 2.84^@"C}}$

The freezing-point depression is defined as the difference bwteen the freezing point of the pure solvent and the freezing point of the solution

$\textcolor{b l u e}{\Delta {T}_{\text{f" = T_"f"^@ - T_"f sol}}}$

This means that the freezing point of the solution will be

${T}_{\text{f sol" = T_"f"^@ - DeltaT_"f}}$

T_"f sol" = 0^@"C" - 2.84^@"C" = color(green)(-2.84^@"C")