Answer
Verified
333.9k+ views
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Fill the blanks with proper collective nouns 1 A of class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Select the word that is correctly spelled a Twelveth class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers