# Question dfd21

Sep 14, 2017

The freezing point is -34 °C.

#### Explanation:

The formula for calculating the freezing point depression ΔT_"f" is

color(blue)(bar(ul(|color(white)(a/a)ΔT_"f" = iK_fbcolor(white)(a/a)|)))" "

where

$i$ is the van't Hoff $i$ factor

${K}_{\text{f}}$ is the molal freezing point depression constant

$b$ is the molality of the solution.

Our first task is to calculate the molality of the solution.

Assume that we mix 1 L of ethylene glycol (EG) and 1 L of water.

$\text{Mass of EG" = 1000 color(red)(cancel(color(black)("mL EG"))) × "1.12 g EG"/(1 color(red)(cancel(color(black)("mL EG")))) = "1120 g EG}$

$\text{Moles of EG" = 1120 color(red)(cancel(color(black)("g EG"))) × "1 mol EG"/(62.07 color(red)(cancel(color(black)("g EG")))) = "18.04 mol EG}$

$\text{Mass of water" = 1000 color(red)(cancel(color(black)("mL water"))) × "1 g water"/(1 color(red)(cancel(color(black)("mL water")))) = "1000 g water" = "1.0 kg water}$

$b = \text{moles solute"/"kilograms solvent" = "18.04 mol"/"1.0 kg" = "18 mol/kg}$

In this problem,

$i = 1$, because EG is a nonelectrolyte
${K}_{\textrm{f}} = \text{1.86 °C·kg·mol"^"-1}$

ΔT_"f" = 1 × "1.86 °C"·color(red)(cancel(color(black)("kg·mol"^"-1"))) × 18.04 color(red)(cancel(color(black)("mol·kg"^"-1"))) = "34 °C"

T_text(f) = T_text(f)^° - ΔT_text(f) = "0 °C - 34 °C" = "-34 °C"#

Note: The answer can have only one significant figure, because that is all you gave for the density of water.

However, I calculated the answer to two significant figures.