What is the change in the freezing point of a solution made by dissolving 14.7g of C₆H₁₂O₆ into 150.0 ml of water? The density of water is and A)0.152°C B)1.01°C C)18.97°C D)1.82°C E)none of the above

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Quizplus · Accepted Answer

The change in freezing point of the solution can be calculated using the formula: ΔTf = Kf x m where Kf is the freezing point depression constant of water (1.86 °C/m), and m is the molality of the solution, which is calculated using the formula: m = moles of solute / mass of solvent (in kg) First, we need to calculate the moles of solute C₆H₁₂O₆: molar mass of C₆H₁₂O₆ = 256 g/mol moles of C₆H₁₂O₆ = 14.7 g / 256 g/mol = 0.0574 mol Next, we need to calculate the mass of solvent: mass of 150.0 ml of water = density x volume = 1.00 g/mL x 150.0 mL = 150.0 g = 0.150 kg Now we can calculate the molality of the solution: m = 0.0574 mol / 0.150 kg = 0.383 mol/kg Finally, we can calculate the ΔTf: ΔTf = Kf x m = 1.86 °C/m x 0.383 mol/kg = 0.71238 °C Rounding to three significant figures gives ΔTf = 0.712 °C, which is closest to choice B (1.01°C).

What Is the Change in the Freezing Point of a Solution