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← IIT-JEE 2008 · Paper 1
ChemistrySingle correct+4 / -1Question 67 of 69

IIT-JEE 2008 · Paper 1, Chemistry question 67

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 freezing point of the solution M is

Choose one option
Answer and worked solution

Official answer: D

Solvent and solute. With mole fraction 0.9, ethanol is the solvent and water the solute: 0.1 mol of water in 0.9×46=41.40.9\times46=41.4 g of ethanol.

Molality. m=0.10.0414=2.42m=\dfrac{0.1}{0.0414}=2.42 mol kg⁻¹.

Freezing point. ΔTf=2.0×2.42=4.83\Delta T_f=2.0\times2.42=4.83 K, so Tf=155.7−4.8=150.9T_f=155.7-4.8=150.9 K.

(D)