Final answer:
To raise the boiling point of water by 9.58 K, 15.54 g of aluminum chloride is needed. This calculation uses the boiling point elevation equation with the ebullioscopic constant of water and the van't Hoff factor for AlCl3, resulting in a molality calculation followed by a conversion to mass.
Step-by-step explanation:
To determine the mass of aluminum chloride needed to raise the boiling point of water by 9.58 K, we use the boiling point elevation equation ΔTb = i * Kb * m, where ΔTb is the boiling point elevation, i is the van't Hoff factor (for AlCl₃, i = 4 since it dissociates into 4 ions: 1 Al³⁺ and 3 Cl⁻), Kb is the ebullioscopic constant of water, and m is the molality of the solution.
First, we calculate the molality (m) needed for the 9.58 K boiling point elevation using the equation:
ΔTb = i * Kb * m => m = ΔTb / (i * Kb)
Plugging in the values we get:
m = 9.58 K / (4 * 0.512 K·kg/mol) = 4.675 mol/kg
To find the mass of AlCl₃ required, we use the following relationship:
molality (m) = moles of solute / mass of solvent in kg
Therefore, moles of AlCl₃ = molality (m) * mass of water in kg
Moles of AlCl₃ = 4.675 mol/kg * 0.025 kg = 0.116875 mol
Finally, to find the mass of AlCl₃:
Mass of AlCl₃ = moles * molar mass = 0.116875 mol * 133 g/mol = 15.54 g