JEE Challenger
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ChemMedAdv 2008 • Paper 1

Freezing Point Depression of Ethanol Water Solution

Comprehension Passage

Paragraph for Question Nos. 67 to 69

Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles

A solution M\mathbf{M} is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.90.9

Given : Freezing point depression constant of water (Kfwater)=1.86 K kg mol−1(\text{K}_\text{f}^\text{water}) = 1.86\text{ K kg mol}^{-1}
Freezing point depression constant of ethanol (Kfethanol)=2.0 K kg mol−1(\text{K}_\text{f}^\text{ethanol}) = 2.0\text{ K kg mol}^{-1}
Boiling point elevation constant of water (Kbwater)=0.52 K kg mol−1(\text{K}_\text{b}^\text{water}) = 0.52\text{ K kg mol}^{-1}
Boiling point elevation constant of ethanol (Kbethanol)=1.2 K kg mol−1(\text{K}_\text{b}^\text{ethanol}) = 1.2\text{ K kg mol}^{-1}
Standard freezing point of water =273 K= 273\text{ K}
Standard freezing point of ethanol =155.7 K= 155.7\text{ K}
Standard boiling point of water =373 K= 373\text{ K}
Standard boiling point of ethanol =351.5 K= 351.5\text{ K}
Vapour pressure of pure water =32.8 mm Hg= 32.8\text{ mm Hg}
Vapour pressure of pure ethanol =40 mm Hg= 40\text{ mm Hg}
Molecular weight of water =18 g mol−1= 18\text{ g mol}^{-1}
Molecular weight of ethanol =46 g mol−1= 46\text{ g mol}^{-1}

In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.

The freezing point of the solution M\mathbf{M} is

Options

A

268.7 K268.7\text{ K}

B

268.5 K268.5\text{ K}

C

234.2 K234.2\text{ K}

D

150.9 K150.9\text{ K}

Correct

Topics & Concepts

SolutionsConcentration TermsMole Concept

Step-by-Step Solution

To determine the freezing point of the solution M\mathbf{M}, we identify the solvent and the solute based on the given mole fractions:

  1. Identification of Solvent and Solute:
    The mole fraction of ethanol is:
    xethanol=0.9x_{\text{ethanol}} = 0.9
    The mole fraction of water is:
    xwater=1−0.9=0.1x_{\text{water}} = 1 - 0.9 = 0.1
    Since ethanol is present in a much larger proportion (90 mol%90\text{ mol}\%), ethanol acts as the solvent and water acts as the solute.

  2. Calculation of Molality of the Solute (Water):
    Consider a sample containing 1 mol1\text{ mol} of the solution:

    • Moles of solute (water), nwater=0.1 moln_{\text{water}} = 0.1\text{ mol}
    • Moles of solvent (ethanol), nethanol=0.9 moln_{\text{ethanol}} = 0.9\text{ mol}

    The mass of the solvent (ethanol) is:
    Wethanol=nethanol×Methanol=0.9 mol×46 g mol−1=41.4 g=0.0414 kgW_{\text{ethanol}} = n_{\text{ethanol}} \times M_{\text{ethanol}} = 0.9\text{ mol} \times 46\text{ g mol}^{-1} = 41.4\text{ g} = 0.0414\text{ kg}

    The molality (mm) of the solution is given by:
    m=nwaterWethanol (in kg)=0.1 mol0.0414 kg≈2.4155 mol kg−1m = \frac{n_{\text{water}}}{W_{\text{ethanol}}\text{ (in kg)}} = \frac{0.1\text{ mol}}{0.0414\text{ kg}} \approx 2.4155\text{ mol kg}^{-1}

  3. Calculation of Depression in Freezing Point (ΔTf\Delta T_f):
    Using the freezing point depression formula:
    ΔTf=Kfethanol×m\Delta T_f = K_f^{\text{ethanol}} \times m
    Given Kfethanol=2.0 K kg mol−1K_f^{\text{ethanol}} = 2.0\text{ K kg mol}^{-1}:
    ΔTf=2.0 K kg mol−1×2.4155 mol kg−1≈4.83 K\Delta T_f = 2.0\text{ K kg mol}^{-1} \times 2.4155\text{ mol kg}^{-1} \approx 4.83\text{ K}

  4. Calculation of the Freezing Point of Solution M\mathbf{M}:
    The standard freezing point of pure ethanol is:
    Tf∘=155.7 KT_f^{\circ} = 155.7\text{ K}

    Therefore, the freezing point of the solution is:
    Tf=Tf∘−ΔTf=155.7 K−4.83 K=150.87 K≈150.9 KT_f = T_f^{\circ} - \Delta T_f = 155.7\text{ K} - 4.83\text{ K} = 150.87\text{ K} \approx 150.9\text{ K}

Thus, the correct option is D.

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