An ethylene glycol solution is mad using 58.5g of (C2H6O2) and diluting to a total volume of 500.0 mL. Calculate the freezing and boiling point. Assume a density of 1.09 g/ml for the solution
To calculate the freezing and boiling point of an ethylene glycol solution, we need to use the formula for freezing point depression and boiling point elevation.
1. Calculate the molality (m) of the ethylene glycol solution:
Molality (m) = moles of solute / mass of solvent (in kg)
First, convert the mass of ethylene glycol (C2H6O2) to moles:
Moles of C2H6O2 = mass / molar mass
= 58.5 g / 62.07 g/mol (molar mass of C2H6O2)
= 0.942 mol
Next, convert the volume of the solution to the mass of the solvent (water):
Mass of water = volume of solution × density of solution
= 500.0 mL × 1.09 g/mL
= 545 g
Finally, calculate the molality:
Molality = 0.942 mol / 0.545 kg
= 1.728 mol/kg
2. Calculate the freezing point depression (∆Tf):
∆Tf = Kf × m
The cryoscopic constant (Kf) for water is approximately 1.86 °C/m.
∆Tf = 1.86 °C/m × 1.728 mol/kg
= 3.209 °C
This means that the freezing point of the ethylene glycol solution will be depressed by 3.209 °C compared to the freezing point of pure water (which is 0 °C).
Freezing point = 0 °C - 3.209 °C
= -3.209 °C
Therefore, the freezing point of the ethylene glycol solution is approximately -3.209 °C.
3. Calculate the boiling point elevation (∆Tb):
∆Tb = Kb × m
The ebullioscopic constant (Kb) for water is approximately 0.512 °C/m.
∆Tb = 0.512 °C/m × 1.728 mol/kg
= 0.8856 °C
This means that the boiling point of the ethylene glycol solution will be elevated by 0.8856 °C compared to the boiling point of pure water (which is 100 °C at sea level).
Boiling point = 100 °C + 0.8856 °C
= 100.8856 °C
Therefore, the boiling point of the ethylene glycol solution is approximately 100.8856 °C.