JEE Advance - Chemistry (2008 - Paper 1 Offline - No. 21)
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 water = 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 water = 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 water = 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.
Explanation
For solution M, water is solute and ethanol is solvent.
$$\because$$ $$\Delta {T_f} = {({K_f})_{ethanol}} \times m$$
$$2 \times {{0.1} \over {0.9 \times {{46} \over {1000}}}} = {{2000} \over {9 \times 46}} = 4.8$$
$$\therefore$$ $${T_f} = T_f^0 - \Delta {T_f} = 155.7 - 4.8 = 150.9\,K$$
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