JEE Advance - Chemistry (2008 - Paper 1 Offline - No. 22)

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 $$\left( {K_f^{water}} \right) = 1.86$$ K kg mol$$^{-1}$$

Freezing point depression constant of ethanol $$\left( {K_f^{ethanol}} \right) = 2.0$$ K kg mol$$^{-1}$$

Boiling point elevation constant of water $$\left( {K_b^{water}} \right) = 0.52$$ K kg mol$$^{-1}$$

Boiling point elevation constant of ethanol $$\left( {K_b^{ethanol}} \right) = 1.2$$ 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.

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 $$\left( {K_f^{water}} \right) = 1.86$$ K kg mol$$^{-1}$$

Freezing point depression constant of ethanol $$\left( {K_f^{ethanol}} \right) = 2.0$$ K kg mol$$^{-1}$$

Boiling point elevation constant of water $$\left( {K_b^{water}} \right) = 0.52$$ K kg mol$$^{-1}$$

Boiling point elevation constant of ethanol $$\left( {K_b^{ethanol}} \right) = 1.2$$ 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.

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 $$\left( {K_f^{water}} \right) = 1.86$$ K kg mol$$^{-1}$$

Freezing point depression constant of ethanol $$\left( {K_f^{ethanol}} \right) = 2.0$$ K kg mol$$^{-1}$$

Boiling point elevation constant of water $$\left( {K_b^{water}} \right) = 0.52$$ K kg mol$$^{-1}$$

Boiling point elevation constant of ethanol $$\left( {K_b^{ethanol}} \right) = 1.2$$ 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 :
39.3 mm Hg
36.0 mm Hg
29.5 mm Hg
28.8 mm Hg

Explanation

Total vapour pressure $$P = P_A^0{x_A}$$

[$$\because$$ Solute is to be taken as non-volatile] $$ = 40 \times 0.9 = 36$$ mm Hg

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