What is the freezing point of a solution containing 0.3 mol of glucose in 1 kg of water? (K_f for water = 1.86 °C kg/mol)

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Q: What is the freezing point of a solution containing 0.3 mol of glucose in 1 kg of water? (K_f for water = 1.86 °C kg/mol)

Solution: Freezing point depression = K_f * m = 1.86 * 0.3 = 0.558 °C; Freezing point = 0 - 0.558 = -0.558 °C

Steps: 8

Step 1: Identify the number of moles of glucose in the solution, which is given as 0.3 mol.
Step 2: Identify the mass of water in the solution, which is given as 1 kg.
Step 3: Find the freezing point depression constant (K_f) for water, which is given as 1.86 °C kg/mol.
Step 4: Calculate the molality (m) of the solution. Since we have 0.3 mol of glucose in 1 kg of water, the molality is 0.3 mol/kg.
Step 5: Use the formula for freezing point depression: Freezing point depression = K_f * m. Substitute the values: 1.86 °C kg/mol * 0.3 mol/kg.
Step 6: Perform the multiplication: 1.86 * 0.3 = 0.558 °C. This is the freezing point depression.
Step 7: Calculate the new freezing point of the solution. The normal freezing point of water is 0 °C, so subtract the freezing point depression from 0 °C: 0 °C - 0.558 °C = -0.558 °C.
Step 8: The final answer is that the freezing point of the solution is -0.558 °C.