Final answer:
The freezing point of a solution containing 5.0 grams of KCl in 550 grams of water is calculated to be approximately -0.454°C, which most closely matches option e) -0.45°C.
Step-by-step explanation:
To calculate the freezing point of a solution containing 5.0 grams of KCl in 550.0 grams of water, we first need to determine the molality of the solution. Potassium chloride (KCl) dissociates into K+ and Cl- ions when dissolved in water. This means that one mole of KCl will produce two moles of particles.
The molar mass of KCl is approximately 74.55 g/mol.
Molality (m) is calculated as the number of moles of solute per kilogram of solvent. First, we find the number of moles of KCl, which is 5.0 g / 74.55 g/mol = 0.0671 mol. Since KCl dissociates into two particles, the number of moles of particles is 0.0671 mol * 2 = 0.1342 mol.
The mass of water in kilograms is 550.0 g or 0.55 kg. The molality of the solution is therefore 0.1342 mol / 0.55 kg = 0.244 mol/kg.
Next, we use the freezing point depression formula: ΔTf = -Kfm, where ΔTf is the freezing point depression, Kf is the molal freezing point depression constant for water (1.86°C kg/mol), and m is the molality of the solution.
Plugging in the values, we get ΔTf = -1.86°C/kg/mol * 0.244 mol/kg = -0.454°C. Thus, the freezing point is lowered by 0.454°C, and since the normal freezing point of water is 0°C, the new freezing point of the solution is about -0.454°C. The closest answer from the options is -0.45°C (option e).