# Question 4f3c7

Jul 9, 2017

WARNING! Long answer! You must add 3.86 L of ethylene glycol. The boiling point of the mixture will be 105.23 °C.

#### Explanation:

(a) Volume of antifreeze

The formula for freezing point depression ΔT_"f" is

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

where

${K}_{\text{f}}$ = the molal freezing point depression constant of the solvent
$b$ = the molality of the solution

We can rearrange the formula to get

b = (ΔT_"f")/K_"f"

ΔT_"f" = "19.0 °C"
${K}_{\text{f" = "1.86 °C·kg·mol"^"-1}}$

b = (19.0 color(red)(cancel(color(black)("°C"))))/(1.86 color(red)(cancel(color(black)("°C")))·"kg·mol"^"-1") = "10.22 mol·kg"^"-1"

$\text{Mass of water" = 6750 color(red)(cancel(color(black)("mL water"))) × "1.00 g water"/(1 color(red)(cancel(color(black)("mL water")))) = "6750 g water" = "6.75 kg water}$

$\text{Amount of EG" = 6.75 color(red)(cancel(color(black)("kg water"))) × "10.22 mol EG"/(1 color(red)(cancel(color(black)("kg water")))) = "68.95 mol EG}$

$\text{Mass of EG" = 68.95 color(red)(cancel(color(black)("mol EG"))) × "62.07 g EG"/(1 color(red)(cancel(color(black)("mol EG")))) = "4280 g EG}$

$\text{Volume of EG" = 4280 color(red)(cancel(color(black)("g EG"))) × "1 mL EG"/(1.11 color(red)(cancel(color(black)("g EG")))) = "3860 mL EG" = "3.86 L EG}$

(b)) Boiling point of solution

The formula for boiling point elevation ΔT_"b" is

color(blue)(bar(ul(|color(white)(a/a)ΔT_"b" = K_"b"bcolor(white)(a/a)|)))" "

where

${K}_{\text{b}}$ = the molal boiling point elevation constant of the solvent ($\text{0.512 °C·kg·mol"^"-1}$).

We can combine the boiling point and freezing point formulas to get

(ΔT_text(b))/(ΔT_text(f)) = (K_"b"color(red)(cancel(color(black)(b))))/(K_text(f)color(red)(cancel(color(black)(b)))) = (K_"b")/(K_text(f))

We can rearrange this equation to get

ΔT_text(b) = ΔT_text(f) × K_text(b)/K_text(f) = "19.0 °C" × (0.512 color(red)(cancel(color(black)("°C·kg·mol"^"-1"))))/(1.86 color(red)(cancel(color(black)("°C·kg·mol"^"-1")))) = "5.23 °C"#

${T}_{\textrm{b}} = \text{100 °C + 5.23 °C = 105.23 °C}$