Final answer:
The boiling point of the solution formed by dissolving 0.593 moles of ethanol in 100.0 grams of water is calculated using molality and the molal boiling point elevation constant. After finding the molality to be 5.93 m, the elevation in boiling point is calculated as 3.04 °C, resulting in a solution boiling point of 103.04 °C. The correct option is (c)
Step-by-step explanation:
To determine the boiling point of the solution formed by dissolving 0.593 moles of ethanol in 100.0 grams of water, we first need to calculate the molality of the solution. Molality (m) is defined as the number of moles of solute (ethanol in this case) per kilogram of solvent (water).
First, we convert grams of water to kilograms:
100.0 g H₂O × (1 kg / 1000 g) = 0.1000 kg H₂O
Next, we use the given number of moles of ethanol and the mass of water to find the molality:
Molality (m) = moles of ethanol / kg of water = 0.593 moles / 0.1000kg = 5.93 m
Now that we have the molality, we can calculate the elevation in boiling point using the formula ΔT = i × Kb × m, where ΔT is the change in boiling point, i is the van't Hoff factor (which is 1 for ethanol because it does not ionize), Kb is the ebullioscopic constant of water, and m is the molality of the solution.
ΔT = 1 × 0.512 °C/m × 5.93 m = 3.04 °C
The boiling point of the solution is then:
BP₀ H₂O + ΔT = 100 °C + 3.04 °C = 103.04 °C
Hence, the boiling point elevation shows that the correct answer for the boiling point of the solution is 103.04 °C.