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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: Water is added to the solution MM such that the mole fraction of water in the solution becomes 0.90.9 The boiling point of this solutions is

Options

  1. A380.4 K380.4 \mathrm{~K}
  2. B376.2 K376.2 \mathrm{~K}
  3. C375.5 K375.5 \mathrm{~K}
  4. D354.7 K354.7 \mathrm{~K}

Correct answer

B. 376.2 K376.2 \mathrm{~K}

Step-by-step solution

χH2O=0.9\chi_{\mathrm{H}_2 \mathrm{O}}=0.9 (solvent) χC2H5OH=0.1\chi_{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}=0.1 (solute) Δ\DeltaT_b=k_b ×\timesm=0.52 ×1.0×10000.9×18=\times \frac{1.0 \times 1000}{0.9 \times 18}=3.2  K\mathrm{~K}Boiling point, Tb=373+3.2=376.2 KT_b=373+3.2=376.2 \mathrm{~K}
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