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"
