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Solutions question

2008 · Shift 1 · Q21
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Solutions question

2008 · Shift 1 · Q21

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 water = 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 freezing point of the solution M is :
  1. A
    268.7 K
  2. B
    268.5 K
  3. C
    234.2 K
  4. D
    150.9 K
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