# Question d7617

Aug 18, 2016

Warning! Long answer. The boiling point is 108.25 °C, and the freezing point is -30.0 °C.

#### Explanation:

Calculate the molality of the solution.

Let's assume that we have 1000 g of the solution of ethylene glycol (EG) in water.

Then

$\text{Mass of EG" = 1000 color(red)(cancel(color(black)("g solution"))) × "25.0 g EG"/(100 color(red)(cancel(color(black)("g solution")))) = "250 g EGr}$

$\text{Mass of water" = "1000 g - 250 g" = "750 g}$

$\text{Moles of EG" = 250color(red)(cancel(color(black)( "g EG"))) × "1 mol EG"/(62.07 color(red)(cancel(color(black)("g EG")))) = "4.028 mol EG}$

$\text{Kilograms of water" = 750 color(red)(cancel(color(black)("g water"))) × "1 kg water"/(1000 color(red)(cancel(color(black)("g water")))) = "0.750 kg water}$

The formula for molality is

color(blue)(|bar(ul(color(white)(a/a) "Molality" = "moles of solute"/"kilograms of solvent"color(white)(a/a)|)))" "

$\text{Molality" = "4.028 mol"/"0.750 kg" = "16.11 mol/kg}$

Boiling point calculation

The formula for boiling point elevation is

color(blue)(|bar(ul(ΔT_"b" = K_"b"m)|)

where

• ΔT_"b" is the boiling point elevation
• ${K}_{\text{b}}$ is the boiling point elevation constant
• $m$ is the molality of the solution

The value of ${K}_{b}$ for water is $\text{0.512 °C·kg"^"-1""mol"^"-1}$.

ΔT_"b" = "0.512 °C·"color(red)(cancel(color(black)("kg·mol"^"-1"))) × 16.11 color(red)(cancel(color(black)("mol·kg"^"-1"))) = "8.25 °C"

T_b = T_b^° + ΔT_b = "100.00 °C + 8.25 °C" = "108.25 °C".

Freezing point calculation

The formula for calculating freezing point depression is

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

where

ΔT_"f" is the decrease in freezing point
${K}_{\text{f}}$ is the molal freezing point depression constant
$\text{m}$ is the molality of the solution.

The molal freezing point depression constant for water is $\text{1.86 °C·kg"^"-1""mol"^"-1}$.

ΔT_f = "1.86 °C"·color(red)(cancel(color(black)("kg"^"-1""mol"^"-1"))) × 16.11 color(red)(cancel(color(black)("mol·kg"^"-1"))) = "30.0 °C"

T_f = T_f^° - ΔT_f = "100.0 °C - 30.0 °C" = "-30.0 °C"#