# Problem: What volume of ethylene glycol (C2H6O2), a nonelectrolyte, must be added to 15.0 L water to produce an antifreeze solution with a freezing point of -25.0°C? What is the boiling point of this solution? (The density of ethylene glycol is 1.11 g/cm3, and the density of water is 1.00 g/cm3.)

🤓 Based on our data, we think this question is relevant for Professor Billman's class at Abilene Christian University.

###### FREE Expert Solution

Step 1: Calculate the freezing point depression of the solution:

$\overline{){\mathbf{∆}}{{\mathbf{T}}}_{{\mathbf{f}}}{\mathbf{=}}{\mathbf{∆}}{\mathbf{T}}{{\mathbf{°}}}_{\mathbf{f}\mathbf{,}\mathbf{pure}}{\mathbf{-}}{\mathbf{∆}}{\mathbf{T}}{{\mathbf{°}}}_{\mathbf{f}\mathbf{,}\mathbf{solution}}}$

${\mathbf{∆}}{{\mathbf{T}}}_{{\mathbf{f}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{°}}{\mathbf{C}}{\mathbf{-}}\left(-25°C\right){\mathbf{=}}$25°C

Step 2: Calculate the molality of the solution:

$\overline{){\mathbf{∆}}{{\mathbf{T}}}_{{\mathbf{f}}}{\mathbf{=}}{{\mathbf{iK}}}_{{\mathbf{f}}}{\mathbf{m}}}$

$\frac{\mathbf{25}\mathbf{°}\mathbf{C}}{\mathbf{\left(}\mathbf{1}\mathbf{\right)}\left(1.86\frac{°C}{m}\right)}{\mathbf{=}}\frac{\overline{)\left(1\right)\left(1.86\frac{°C}{m}\right)}\left(m\right)}{\overline{)\mathbf{\left(}\mathbf{1}\mathbf{\right)}\left(1.86\frac{°C}{m}\right)}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathbit{m}}{\mathbf{=}}\frac{\mathbf{25}\overline{)\mathbf{°}\mathbf{C}}}{\mathbf{\left(}\mathbf{1}\mathbf{\right)}\left(1.86\frac{\overline{)°C}}{m}\right)}$

###### Problem Details

What volume of ethylene glycol (C2H6O2), a nonelectrolyte, must be added to 15.0 L water to produce an antifreeze solution with a freezing point of -25.0°C? What is the boiling point of this solution? (The density of ethylene glycol is 1.11 g/cm3, and the density of water is 1.00 g/cm3.)