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JEE Advanced 2008Paper 1
Exam
JEE Advanced
Year
2008
Paper
Paper 1
Subject
Chemistry
Chapter
Solutions

JEE Advanced 2008 · Paper 1 · Chemistry · Solutions

Paragraph: Properties such as boiling point, freezing point and...

Paragraph: 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 automobiles. A solution MM is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixtrue is 0.90.9 Given Freezing point depression constant of water (kwater )=1.86 K kg mol−1\left(k^{\text{water }}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} Freezing point depression constant of ethanol (kfethanol )=2.0Kkgmol−1\left(k_f^{\text{ethanol }}\right)=2.0 \mathrm{Kkg} \mathrm{mol}^{-1} Boiling point elevation constant of water (kbwater )=0.52Kkgmol−1\left(k_b^{\text{water }}\right)=0.52 \mathrm{Kkg} \mathrm{mol}^{-1} Boiling point elevation constant of ethanol (kbethanol )=1.2Kkgmol−1\left(k_b^{\text{ethanol }}\right)=1.2 \mathrm{Kkg} \mathrm{mol}^{-1} Standard freezing point of water =273 K=273 \mathrm{~K} Standard freezing point of ethanol =155.7 K=155.7 \mathrm{~K} Standard boiling point of water =373 K=373 \mathrm{~K} Standard boiling point of ethanol =351.5 K=351.5 \mathrm{~K} Vapour pressure of pure water =328 mmofHg=328 \mathrm{~mm} of \mathrm{Hg} Vapour pressure of pure ethanol =40 mmofHg=40 \mathrm{~mm} of \mathrm{Hg} Molecular weight of water =18 g mol−1=18 \mathrm{~g} \mathrm{~mol}^{-1} Molecular weight of ethanol =46 g mol−1=46 \mathrm{~g} \mathrm{~mol}^{-1} In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.Question: The vapour pressure of the solution MM is

Options

  1. A39.3 mmHg39.3 \mathrm{~mm} \mathrm{Hg}
  2. B36.0 mmHg36.0 \mathrm{~mm} \mathrm{Hg}
  3. C29.5 mmHg29.5 \mathrm{~mm} \mathrm{Hg}
  4. D28.8 mmHg28.8 \mathrm{~mm} \mathrm{Hg}

Correct answer

B. 36.0 mmHg36.0 \mathrm{~mm} \mathrm{Hg}

Step-by-step solution

Total vapour pressure, p=pA∘χAp=p_A^{\circ} \chi_A p=40 \times 0.9=36 \mathrm{ mm} \text{ of } Hg\mathrm{Hg} In the paragraph, it has been directed to take solute as non-volatile, thus H2O\mathrm{H}_2 \mathrm{O} do not contribute in the total vapour pressure.
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