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Hint: We should know that Raoult's law is related to vapour pressure. You must also know the meaning of what volatile is, easy evaporation of a substance at normal temperatures. Now you can easily answer this question.
Complete step by step answer:
Raoult's law - The vapour pressure of a solution of a non-volatile solute is equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction.
Mathematically, Raoult’s law equation is written as;
$P_{ solution }\quad =\quad P^{ 0 }_{ solvent }X_{ solvent }$
In this equation, $P^{ 0 }_{ solvent }$ is the vapour pressure of the pure solvent at a particular temperature, $P_{ solution }$ is the total pressure of the solution, $X_{ solvent }$ is the mole fraction of the solvent.
Proof of Raoult's law for non-volatile solute in volatile solvent -
For a solution of a volatile solute (A) and solvent (B) total pressure is sum of the partial pressure of the two as:
$P_{ solution }\quad =\quad P^{ 0 }_{ A }X_{ A }\quad +\quad P^{ 0 }_{ B }X_{ B }$
As we know $X_{ A }$ + $X_{ B }$ = 1 , Or we can write this as $X_{ A }$ = 1 - $X_{ B }$
Now place this value in above equation,
$P_{ solution }\quad =\quad P^{ 0 }_{ A }(1-X_{ B })\quad +\quad P^{ 0 }_{ B }X_{ B }$
Now rearrange this equation,
$P_{ solution }\quad =\quad P^{ 0 }_{ A }\quad +\quad (P^{ 0 }_{ B }-P^{ 0 }_{ A })X_{ B }$
We know that, for non-volatile solute $P^{ 0 }_{ A }$ = 0
Thus, $P_{ solution }\quad =\quad P^{ 0 }_{ B }X_{ B }$
Limitations of Raoult's law:
Intermolecular forces between the solvent and solute components should be similar to those between individual molecules.
The gaseous phase is assumed to behave ideal, where ideal gas law can be applied.
Therefore, we stated and proved Raoult's law for non-volatile solute in volatile solvent with the limitations of the law.
Note: We can do a comparison between Raoult's law and Henry's law to understand the difference. Henry's law is a limiting law that only applies for "sufficiently dilute" solutions, while Raoult's law is generally valid when the liquid phase is almost pure or for mixtures of similar substances.
Complete step by step answer:
Raoult's law - The vapour pressure of a solution of a non-volatile solute is equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction.
Mathematically, Raoult’s law equation is written as;
$P_{ solution }\quad =\quad P^{ 0 }_{ solvent }X_{ solvent }$
In this equation, $P^{ 0 }_{ solvent }$ is the vapour pressure of the pure solvent at a particular temperature, $P_{ solution }$ is the total pressure of the solution, $X_{ solvent }$ is the mole fraction of the solvent.
Proof of Raoult's law for non-volatile solute in volatile solvent -
For a solution of a volatile solute (A) and solvent (B) total pressure is sum of the partial pressure of the two as:
$P_{ solution }\quad =\quad P^{ 0 }_{ A }X_{ A }\quad +\quad P^{ 0 }_{ B }X_{ B }$
As we know $X_{ A }$ + $X_{ B }$ = 1 , Or we can write this as $X_{ A }$ = 1 - $X_{ B }$
Now place this value in above equation,
$P_{ solution }\quad =\quad P^{ 0 }_{ A }(1-X_{ B })\quad +\quad P^{ 0 }_{ B }X_{ B }$
Now rearrange this equation,
$P_{ solution }\quad =\quad P^{ 0 }_{ A }\quad +\quad (P^{ 0 }_{ B }-P^{ 0 }_{ A })X_{ B }$
We know that, for non-volatile solute $P^{ 0 }_{ A }$ = 0
Thus, $P_{ solution }\quad =\quad P^{ 0 }_{ B }X_{ B }$
Limitations of Raoult's law:
Intermolecular forces between the solvent and solute components should be similar to those between individual molecules.
The gaseous phase is assumed to behave ideal, where ideal gas law can be applied.
Therefore, we stated and proved Raoult's law for non-volatile solute in volatile solvent with the limitations of the law.
Note: We can do a comparison between Raoult's law and Henry's law to understand the difference. Henry's law is a limiting law that only applies for "sufficiently dilute" solutions, while Raoult's law is generally valid when the liquid phase is almost pure or for mixtures of similar substances.
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