Final answer:
The freezing point of 0.312 kg of ethanol containing 15.84 g of ethylene glycol is calculated to be approximately -116.22 °C after determining the moles of solute, the molality of the solution, and applying the freezing point depression equation.
Step-by-step explanation:
To calculate the freezing point of 0.312 kg of ethanol containing 15.84 g of ethylene glycol (C2H6O2), we need to follow a series of steps involving colligative properties. First, we determine the moles of ethylene glycol by dividing its mass by its molar mass:
Number of moles = mass (g) / molar mass (g/mol)
The molar mass of ethylene glycol (C2H6O2) is approximately 62.07 g/mol.
Thus, the number of moles of ethylene glycol in our solution is:
15.84 g / 62.07 g/mol ≈ 0.255 mol
Next, we calculate the molality (m) of the ethylene glycol in ethanol using the formula:
Molality (m) = moles of solute / mass of solvent (kg)
0.255 mol / 0.312 kg ≈ 0.817 mol/kg
Now we use the freezing point depression equation, ΔTf = Kf × m, where ΔTf is the freezing point depression, Kf is the freezing point depression constant for ethanol, and m is the molality calculated above.
With the given Kf for ethanol of 1.99 °C·kg/mol, we find the freezing point depression:
ΔTf = 1.99 °C·kg/mol × 0.817 mol/kg ≈ 1.62 °C
Lastly, we subtract the ΔTf from the normal freezing point of pure ethanol to find the freezing point of the solution:
Freezing point of solution = normal freezing point of ethanol - ΔTf
Freezing point of solution = -114.6 °C - 1.62 °C ≈ -116.22 °C
Therefore, the freezing point of 0.312 kg of ethanol containing 15.84 g of ethylene glycol is approximately -116.22 °C.