Question:

A solution is prepared by dissolving 10 g of a non-volatile solute in 100 g of water. The freezing point depression of the solution is observed to be 1.5°C. Calculate the cryoscopic constant of the solvent.

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Cryoscopic constant (KfK_f) is a measure of the solvent’s ability to decrease its freezing point when a solute is added. Use the formula ΔTf=Kf×m\Delta T_f = K_f \times m to calculate it.
Updated On: Jan 17, 2025
  • 0.15kg/mol0.15 \, \text{kg/mol}
  • 1.5kg/mol1.5 \, \text{kg/mol}
  • 5.0kg/mol5.0 \, \text{kg/mol}
  • 2.0kg/mol2.0 \, \text{kg/mol}
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The Correct Option is A

Solution and Explanation

The depression in freezing point (ΔTf\Delta T_f) is related to the cryoscopic constant (KfK_f) by the formula: ΔTf=Kf×m, \Delta T_f = K_f \times m, where mm is the molality of the solution, and KfK_f is the cryoscopic constant of the solvent. To calculate molality (mm), we use the formula: m=moles of solutemass of solvent in kg. m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}. First, calculate the moles of the solute: moles of solute=mass of solutemolar mass of solute. \text{moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}}. We are not provided with the molar mass of the solute, so we assume it is 10 g/mol for simplicity. Therefore, the moles of solute is: moles of solute=1010=1mol. \text{moles of solute} = \frac{10}{10} = 1 \, \text{mol}. Now, calculate the molality: m=1mol0.1kg=10mol/kg. m = \frac{1 \, \text{mol}}{0.1 \, \text{kg}} = 10 \, \text{mol/kg}. Substitute this value into the depression in freezing point formula: 1.5=Kf×10Kf=1.510=0.15kg/mol. 1.5 = K_f \times 10 \quad \Rightarrow \quad K_f = \frac{1.5}{10} = 0.15 \, \text{kg/mol}. Thus, the cryoscopic constant KfK_f of the solvent is 0.15kg/mol0.15 \, \text{kg/mol}.
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