We’re being asked to calculate for the freezing point of an aqueous solution containing 50% (v/v) ethylene glycol

Recall that the freezing point of a solution is lower than that of the pure solvent and the ** change in freezing point (ΔT_{f})** is given by:

$\overline{){{\mathbf{\Delta T}}}_{{\mathbf{f}}}{\mathbf{=}}{{\mathbf{\Delta T}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{solution}}{\mathbf{-}}{{\mathbf{\Delta T}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{pure}\mathbf{}\mathbf{solvent}}}$

The ** change in freezing point** is also related to the molality of the solution:

$\overline{){{\mathbf{\Delta T}}}_{{\mathbf{f}}}{\mathbf{=}}{\mathbf{i}}{\mathbf{}}{{\mathbf{K}}}_{{\mathrm{f}}}{\mathbf{}}{\mathbf{m}}}$

where:

**i** = van’t Hoff factor

**m** = molality of the solution (in m or mol/kg)

**K _{f}** = freezing point depression constant (in ˚C/m)

We need to convert the 50% (v/v) of ethylene glycol to molality. Recall that the ** molality of a solution** is given by:

$\overline{){\mathbf{Molality}}{\mathbf{}}{\mathbf{\left(}}{\mathbf{m}}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solute}}{\mathbf{Kilograms}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solvent}}}$

**We will calculate the freezing point of the solution using the following steps:**

**Step 1**. Determine the **composition** of the solution.**Step 2**. Calculate the **moles of the solute**.**Step 3**. Calculate the **mass of the solvent** (in kg).**Step 4**. Calculate the **molality of the solution**.**Step 5**. Calculate the** freezing point of the solution**

**Step 1**. Determine the **composition** of the solution.

Radiator coolant is often a 50/50 % by volume mixture of ethylene glycol, HOCH2CH2OH (62.1 g/mol), and water. At 20°C, the density of ethylene glycol is 1.1088 g/mL and the density of water is 0.9982 g/mL. Assuming that the volumes are additive, what is the expected freezing point (°C) of a 50/50(v/v)% ethylene glycol/water solution? Kf = 1.86°C/m for water. Enter your answer as the nearest whole number with no units.

a. 100.53°C

b. 101.23°C

c. 100.08°C

d. 103.91°C

e. 100.31°C

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