
Which of the following solutions is an example of a negative deviation from Raoult’s Law?
A.Acetone $ + $ Ethanol
B.Carbon tetrachloride $ + $ chloroform
C.Acetone $ + $ Chloroform
D.Water $ + $ ethanol
Answer
558.9k+ views
Hint: To answer this question, recall the concept of Raoult’s law. According to this law a solvent’s partial vapour pressure in a solution (or mixture) is equal or identical to the vapour pressure of the pure solvent multiplied by its mole fraction in the solution.
Complete step by step solution:
The mathematical expression of Raoult’s law can be written as:
\[{\text{P}}_{\text{solution}} = \;{\rm{X}}_{\text{solvent}}{\text{P}}^{\text{0}}_{\text{solvent}}\]
where, \[{{\text{P}}_{{\text{solution}}}}\] = vapour pressure of the solution, \[{{{{\rm X}}}_{{\text{solvent}}}}{\text{}}\] = mole fraction of the solvent, \[{{\text{P}}^{\text{0}}}_{{\text{solvent}}}\] = vapour pressure of the pure solvent. The negative and positive deviation is present in solutions called Non-ideal solutions which have
The solute-solute and solvent-solvent interaction are different from the interaction between solute and solvent.
The enthalpy of mixing \[{\Delta _{{\text{mix}}}}{\text{H}} \ne 0\], this implies that heat might be released \[\left( {{\Delta _{{\text{mix}}}}{\text{H}} < 0} \right)\]or the heat might have been absorbed \[\left( {{\Delta _{{\text{mix}}}}{\text{H}} > 0} \right)\]
The volume of mixing \[\left( {{\Delta _{{\text{mix}}}}{\text{V}} \ne 0} \right)\] which means that that there will be some expansion or contraction in the dissolution of liquids
Negative Deviation occurs when the total vapour pressure is less than what it should be according to Raoult’s Law. Considering two components in a solution A and B forming a non-ideal solution, the features it will show are:
\[{{\text{P}}_{\text{A}}} < {{\text{P}}_{\text{A}}}^{\text{0}}{\text{}}{x_A}\;\] and \[{{\text{P}}_{\text{B}}} < {{\text{P}}^{\text{0}}}_{\text{B}}\;{x_B}\;\]as the total vapour pressure \[\left( {{{\text{P}}_{\text{A}}}^{\text{0}}{x_A} + {{\text{P}}^{\text{0}}}_{\text{B}}{x_B}} \right)\] is less than what it should be concerning Raoult’s Law
The solute-solvent interaction is stronger than solute-solute and solvent-solvent interaction that is, \[A-B{\text{ }} > {\text{ }}A-A{\text{ or }}B-B\]
The enthalpy of mixing is negative \[{\Delta _{{\text{mix}}}}{\text{H}} < 0\]
The volume of mixing is negative that is, \[{\Delta _{{\text{mix}}}}{\text{V}} < 0\]
Amongst all the options acetone and chloroform form a non-ideal solution showing negative deviation from Raoult's law.
Therefore, we can conclude that the correct answer to this question is option C.
Note:
You should know about the Limitations of Raoult’s Law. In practical life, ideal solutions are hard to find as the components have to be chemically identical equally. Since many of the components of the solution mixture are not in terms of attractive forces, these types of solutions tend to deviate away from the law.
Complete step by step solution:
The mathematical expression of Raoult’s law can be written as:
\[{\text{P}}_{\text{solution}} = \;{\rm{X}}_{\text{solvent}}{\text{P}}^{\text{0}}_{\text{solvent}}\]
where, \[{{\text{P}}_{{\text{solution}}}}\] = vapour pressure of the solution, \[{{{{\rm X}}}_{{\text{solvent}}}}{\text{}}\] = mole fraction of the solvent, \[{{\text{P}}^{\text{0}}}_{{\text{solvent}}}\] = vapour pressure of the pure solvent. The negative and positive deviation is present in solutions called Non-ideal solutions which have
The solute-solute and solvent-solvent interaction are different from the interaction between solute and solvent.
The enthalpy of mixing \[{\Delta _{{\text{mix}}}}{\text{H}} \ne 0\], this implies that heat might be released \[\left( {{\Delta _{{\text{mix}}}}{\text{H}} < 0} \right)\]or the heat might have been absorbed \[\left( {{\Delta _{{\text{mix}}}}{\text{H}} > 0} \right)\]
The volume of mixing \[\left( {{\Delta _{{\text{mix}}}}{\text{V}} \ne 0} \right)\] which means that that there will be some expansion or contraction in the dissolution of liquids
Negative Deviation occurs when the total vapour pressure is less than what it should be according to Raoult’s Law. Considering two components in a solution A and B forming a non-ideal solution, the features it will show are:
\[{{\text{P}}_{\text{A}}} < {{\text{P}}_{\text{A}}}^{\text{0}}{\text{}}{x_A}\;\] and \[{{\text{P}}_{\text{B}}} < {{\text{P}}^{\text{0}}}_{\text{B}}\;{x_B}\;\]as the total vapour pressure \[\left( {{{\text{P}}_{\text{A}}}^{\text{0}}{x_A} + {{\text{P}}^{\text{0}}}_{\text{B}}{x_B}} \right)\] is less than what it should be concerning Raoult’s Law
The solute-solvent interaction is stronger than solute-solute and solvent-solvent interaction that is, \[A-B{\text{ }} > {\text{ }}A-A{\text{ or }}B-B\]
The enthalpy of mixing is negative \[{\Delta _{{\text{mix}}}}{\text{H}} < 0\]
The volume of mixing is negative that is, \[{\Delta _{{\text{mix}}}}{\text{V}} < 0\]
Amongst all the options acetone and chloroform form a non-ideal solution showing negative deviation from Raoult's law.
Therefore, we can conclude that the correct answer to this question is option C.
Note:
You should know about the Limitations of Raoult’s Law. In practical life, ideal solutions are hard to find as the components have to be chemically identical equally. Since many of the components of the solution mixture are not in terms of attractive forces, these types of solutions tend to deviate away from the law.
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