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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 freezing point of the solution MM is

Options

  1. A268.7 K268.7 \mathrm{~K}
  2. B268.5 K268.5 \mathrm{~K}
  3. C234.2 K234.2 \mathrm{~K}
  4. D150.9 K150.9 \mathrm{~K}

Correct answer

D. 150.9 K150.9 \mathrm{~K}

Step-by-step solution

Solution MM is mixture of ethanol and water. Mole fraction of ethanol is 0.9⇒0.9 \Rightarrow Solvent is C2H5OH\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} Mole fraction of water is 0.1⇒H2O0.1 \Rightarrow \mathrm{H}_2 \mathrm{O} is solute Molality of H2O=n2n1M1=0.10.9×46×1000=2.415Δ\mathrm{H}_2 \mathrm{O}=\frac{n_2}{n_1 M_1}=\frac{0.1}{0.9 \times 46} \times 1000=2.415\DeltaT_f=k_f ⋅\cdotm=2 ×\times2.415=4.83 or Freezing point of solution =155.7−4.83=150.87 K=155.7-4.83=150.87 \mathrm{~K}
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