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Solutions question

2008 · Shift 1 · Q22
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Solutions question

2008 · Shift 1 · Q22

JEE AdvancedChemistrySolutionsMCQ+3 / −1
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 is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9. Given: Freezing point depression constant of water (Kfwater)=1.86\left( {K_f^{water}} \right) = 1.86(Kfwater​)=1.86 K kg mol −1{}^{-1}−1 Freezing point depression constant of ethanol (Kfethanol)=2.0\left( {K_f^{ethanol}} \right) = 2.0(Kfethanol​)=2.0 K kg mol −1{}^{-1}−1 Boiling point elevation constant of water (Kbwater)=0.52\left( {K_b^{water}} \right) = 0.52(Kbwater​)=0.52 K kg mol −1{}^{-1}−1 Boiling point elevation constant of ethanol (Kbethanol)=1.2\left( {K_b^{ethanol}} \right) = 1.2(Kbethanol​)=1.2 K kg mol −1{}^{-1}−1 Standard freezing point of water = 273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecular weight of water = 18 g mol −1{}^{-1}−1 Molecular weight of ethanol = 46 g mol −1{}^{-1}−1 In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.The vapour pressure of the solution M is :
  1. A
    39.3 mm Hg
  2. B
    36.0 mm Hg
  3. C
    29.5 mm Hg
  4. D
    28.8 mm Hg
View written solutionFree

Correct answer: B

Step-by-Step Solution:

  1. Identify the Components and Composition of the Solution: The solution M is prepared by mixing ethanol and water. The mole fraction of ethanol, xethanolx_{ethanol}xethanol​, is given as 0.9. Since the sum of mole fractions in a binary mixture is 1, the mole fraction of water, xwaterx_{water}xwater​, can be calculated as: xwater=1−xethanol=1−0.9=0.1x_{water} = 1 - x_{ethanol} = 1 - 0.9 = 0.1xwater​=1−xethanol​=1−0.9=0.1

  2. Identify the Solvent and Solute: In a solution, the component with the higher mole fraction is considered the solvent, and the one with the lower mole fraction is the solute. Since xethanol=0.9x_{ethanol} = 0.9xethanol​=0.9 and xwater=0.1x_{water} = 0.1xwater​=0.1, ethanol is the solvent and water is the solute.

  3. Apply Raoult's Law with the Given Conditions: The problem states: "In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative." This is a crucial instruction. Although water is physically a volatile substance, for this specific problem, we must treat it as a non-volatile solute because it is the component with the lower mole fraction.

  4. Formulate the Vapour Pressure Equation: According to Raoult's law for a solution containing a non-volatile solute, the vapour pressure of the solution is solely due to the solvent. The total vapour pressure (PsolutionP_{solution}Psolution​) is equal to the partial pressure of the solvent. Psolution=Psolvent=Psolvent0×xsolventP_{solution} = P_{solvent} = P^0_{solvent} \times x_{solvent}Psolution​=Psolvent​=Psolvent0​×xsolvent​ In this case, the solvent is ethanol. Psolution=Pethanol0×xethanolP_{solution} = P^0_{ethanol} \times x_{ethanol}Psolution​=Pethanol0​×xethanol​

  5. Substitute the Given Values and Calculate: From the problem statement:

    • Vapour pressure of pure ethanol, Pethanol0=40P^0_{ethanol} = 40Pethanol0​=40 mm Hg
    • Mole fraction of ethanol, xethanol=0.9x_{ethanol} = 0.9xethanol​=0.9

    Substituting these values into the equation: Psolution=40 mm Hg×0.9P_{solution} = 40 \text{ mm Hg} \times 0.9Psolution​=40 mm Hg×0.9 Psolution=36 mm HgP_{solution} = 36 \text{ mm Hg}Psolution​=36 mm Hg

    Note that the vapour pressure of pure water (Pwater0=32.8P^0_{water} = 32.8Pwater0​=32.8 mm Hg) is extraneous information based on the specific instruction to treat the solute as non-volatile.

  6. Compare with Options: The calculated vapour pressure of the solution is 36.0 mm Hg. This matches option B.

Conclusion:

Based on the strict interpretation of the problem's instruction to treat the solute as non-volatile, the vapour pressure of the solution is 36.0 mm Hg.

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