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  • Boiling & Freezing Point Changes with Glucose Solution: A Calculation
    Here's how to calculate the change in boiling point and freezing point of water when 45 grams of glucose (C6H12O6) are dissolved in 500 grams of water:

    Understanding the Concepts

    * Boiling Point Elevation: When a solute (like glucose) is dissolved in a solvent (like water), the boiling point of the solution increases compared to the pure solvent.

    * Freezing Point Depression: Similarly, the freezing point of the solution decreases compared to the pure solvent.

    * Colligative Properties: These changes in boiling and freezing points are examples of colligative properties, which depend only on the concentration of solute particles, not their identity.

    Calculations

    1. Moles of Glucose:

    * Molar mass of glucose (C6H12O6) = 180.156 g/mol

    * Moles of glucose = 45 g / 180.156 g/mol ≈ 0.25 mol

    2. Molality (m):

    * Molality is moles of solute per kilogram of solvent.

    * Mass of water (solvent) = 500 g = 0.5 kg

    * Molality (m) = 0.25 mol / 0.5 kg = 0.5 mol/kg

    3. Boiling Point Elevation (ΔTb):

    * ΔTb = Kb * m

    * Kb (ebullioscopic constant for water) = 0.512 °C/m

    * ΔTb = 0.512 °C/m * 0.5 mol/kg = 0.256 °C

    4. Freezing Point Depression (ΔTf):

    * ΔTf = Kf * m

    * Kf (cryoscopic constant for water) = 1.86 °C/m

    * ΔTf = 1.86 °C/m * 0.5 mol/kg = 0.93 °C

    Results:

    * Change in Boiling Point: The boiling point of the water will increase by 0.256 °C.

    * Change in Freezing Point: The freezing point of the water will decrease by 0.93 °C.

    Important Note: These calculations assume that glucose does not dissociate in water. If the solute were an ionic compound (like NaCl), it would dissociate into multiple ions, leading to a larger change in boiling and freezing points.

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