(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_"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"
In your problem,
ΔT_"f" = "19.0 °C"
K_"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"
"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"
"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"
"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"
"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"
You must add 3.86 L of ethylene glycol to the radiator.
(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_"b" = the molal boiling point elevation constant of the solvent ("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_text(b) = "100 °C + 5.23 °C = 105.23 °C"
