JEE Advance - Chemistry (2008 - Paper 1 Offline - No. 23)
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
$$\matrix{ {\mathop {{x_{ethanol}} = 0.1}\limits_{(\mathrm{Solute \equiv B})} } & : & {\mathop {{x_{water}} = 0.9}\limits_{(\mathrm{Solvent \equiv A})} } \cr } $$
$$\Delta {T_b} = {K_b}\,.\,m = {K_b}\,.\,{{{x_B}} \over {1 - {x_B}}} \times {{1000} \over {{M_{{w_A}}}}}$$
$$ = 0.52 \times {{0.1} \over {0.9}} \times {{1000} \over {18}} = 3.2\,K$$
$${T_b} = T_b^0 + \Delta {T_b} = 373 + 3.2 = 376.2\,K$$
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