What is the freezing point of the solution with 37 g of MgCl2 dissolved in 0.48 kg of ethanol? Kf ethanol = 2.00°C/m; Mg= 24.31 g/mol; Cl= 35.45 g/mol
Answer (1)
To find the freezing point of the solution, we'll use the formula:ΔTf = Kf × mwhere:ΔTf = freezing point depression (°C)Kf = cryoscopic constant (°C/m) = 2.00°C/mm = molality (moles of solute/kg of solvent)First, calculate the number of moles of MgCl2:moles MgCl2 = mass MgCl2 / molar mass MgCl2moles MgCl2 = 37 g / (24.31 g/mol + 2 × 35.45 g/mol)moles MgCl2 = 37 g / 95.21 g/molmoles MgCl2 ≈ 0.388 molNow, calculate the molality:m = moles MgCl2 / mass ethanol (kg)m = 0.388 mol / 0.48 kgm ≈ 0.808 mNext, calculate the freezing point depression:ΔTf = Kf × mΔTf = 2.00°C/m × 0.808 mΔTf ≈ 1.616°CThe freezing point of pure ethanol is 114.14°C (or -114.14°C, checking for the unit). To find the freezing point of the solution:Tf (solution) = Tf (pure ethanol) - ΔTfTf (solution) = -114.14°C - 1.616°CTf (solution) ≈ -115.756°CSo, the freezing point of the solution is approximately -115.76°C.Note: The negative sign indicates that the freezing point is below 0°C.
