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_"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.
#"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"#
#"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"#
#"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 = "moles solute"/"kilograms solvent" = "18.04 mol"/"1.0 kg" = "18 mol/kg"#
In this problem,
#i= 1#, because EG is a nonelectrolyte
#K_text(f) = "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.
