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← IIT-JEE 2008 · Paper 1
ChemistrySingle correct+4 / -1Question 68 of 69

IIT-JEE 2008 · Paper 1, Chemistry question 68

Passage

Paragraph for Question Nos. 67 to 69

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(K_f^{water})=1.86 K kg mol⁻¹
  • Freezing point depression constant of ethanol (Kfethanol)=2.0(K_f^{ethanol})=2.0 K kg mol⁻¹
  • Boiling point elevation constant of water (Kbwater)=0.52(K_b^{water})=0.52 K kg mol⁻¹
  • Boiling point elevation constant of ethanol (Kbethanol)=1.2(K_b^{ethanol})=1.2 K kg mol⁻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
  • Molecular weight of ethanol = 46 g mol⁻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

Choose one option
Answer and worked solution

Official answer: B

Raoult's law with the solute treated as non-volatile (as the passage instructs): only the solvent, ethanol, contributes:
p=xethanol pethanol∘=0.9×40=36.0 mm Hg.p=x_\text{ethanol}\,p^\circ_\text{ethanol}=0.9\times40=36.0\text{ mm Hg}.

(B)