Final answer:
The freezing point of a solution with 651 g of ethylene glycol in 2505 g of water is -7.786 °C, and its boiling point is 102.17672 °C, determined by using the molality of the solution and the freezing point depression and boiling point elevation equations.
Step-by-step explanation:
To calculate the freezing point of a solution containing 651 g of ethylene glycol in 2505 g of water, you need to first determine the molality of the solution using the freezing point depression constant (Kf). The molecular weight of ethylene glycol (C2H6O2) is approximately 62.07 g/mol. Divide the mass of ethylene glycol by its molecular weight to find the moles of solute.
- moles of ethylene glycol = 651 g / 62.07 g/mol = 10.488 mol
Then, calculate the molality (m) by dividing the moles of solute by the kilograms of solvent (water).
- molality (m) = 10.488 mol / 2.505 kg = 4.186 m
Now, apply the freezing point depression equation: ΔTf = i * Kf * m, where i is the van't Hoff factor (which is 1 for ethylene glycol, as it does not dissociate in solution).
- ΔTf = 1 * 1.86 °C/m * 4.186 m = 7.786 °C
The freezing point of the solution is calculated by subtracting the depression from the normal freezing point of water, 0 °C.
- Freezing point of solution = 0 °C - 7.786 °C = -7.786 °C
To determine the boiling point, use the boiling point elevation equation ΔTb = i * Kb * m. Since the normal boiling point of water is 100 °C and the boiling point elevation constant (Kb) is given as 0.52 °C/m, you can calculate:
- ΔTb = 1 * 0.52 °C/m * 4.186 m = 2.17672 °C
- Boiling point of solution = 100 °C + 2.17672 °C = 102.17672 °C
The boiling point of the ethylene glycol solution is 102.17672 °C.