**Calculate the molality of the solution.**

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

Then

#"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"#

#"Mass of water" = "1000 g - 250 g" = "750 g"#

#"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"#

#"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)|)))" "#

∴ #"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_"b"# is the boiling point elevation constant
- #m# is the molality of the solution

The value of #K_b# for water is #"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_"f"# is the molal freezing point depression constant

#"m"# is the molality of the solution.

The molal freezing point depression constant for water is #"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"#