Chemistry, asked by AlokTyagi, 8 months ago

State Raoult’s law. Non-ideal solutions exhibit either positive or negative deviations from Raoult’s law. What are these deviations and why are they caused? Explain with one example for each type. Derive an expression for Raoult’s law when the solute is non-volatile. ​

Answers

Answered by XxItzkillergirlXx
7

 \huge{ \underline{  \underline{\mathbb{ \red{A}{ \pink{N}{ \blue{S}{ \purple{W}{ \orange{E}{ \green{R} - }}}}}}}}}

 \underline{ \tt{ \red{What  \: is  \: Raoult’s  \: Law?}}}

Raoult’s law states that 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.

Mathematically, Raoult’s law equation is written as;

Psolution = ΧsolventP0solvent

Where,

Psolution = vapour pressure of the solution

Χsolvent = mole fraction of the solvent

P0solvent = vapour pressure of the pure solvent

We will further understand the principle behind the law by looking at the example below.

Consider a solution of volatile liquids A and B in a container. Because A and B are both volatile, there would be both particles of A and B in the vapour phase.

Hence, the vapour particles of both A and B exert partial pressure which contributes to the total pressure above the solution.

Solution of volatile liquids A and B

Raoult’s Law further states that at equilibrium,

{{P}_{A}}=P_{A}^{{}^\circ }{{x}_{A}},{{P}_{B}}=P_{B}^{{}^\circ }{{x}_{B}}P

A

=P

A

x

A

,P

B

=P

B

x

B

Where PA is the partial pressure of A.

P_{A}^{{}^\circ }P

A

is vapour pressure of pure A at that temperature.

{{x}_{A}}x

A

is mole fraction of A in the liquid phase.

Similarly {{P}_{B}},P_{B}^{{}^\circ }{{x}_{B}}P

B

,P

B

x

B

Hence {{P}_{T}}={{P}_{A}}+{{P}_{B}}\left( Dalton’s\,Law \right)P

T

=P

A

+P

B

(Dalton’sLaw) =P_{A}^{{}^\circ }{{x}_{A}}+P_{B}^{{}^\circ }{{x}_{B}}=P

A

x

A

+P

B

x

B

=P_{A}^{{}^\circ }+{{x}_{B}}\left( P_{B}^{{}^\circ }-P_{A}^{{}^\circ } \right)=P

A

+x

B

(P

B

−P

A

)

{ \underline{ \tt{ \red{Limitations \:  of  \: Raoult’s \:  Law}}}}

There are a few limitations to Raoult’s law.

Raoult’s law is apt for describing ideal solutions. However, ideal solutions are hard to find and they are rare. Different chemical components have to be chemically identical equally.

Since many of the liquids that are in the mixture do not have the same uniformity in terms of attractive forces, these type of solutions tends to deviate away from the law.

There is either a negative or a positive deviation. The negative deviation occurs when the vapour pressure is lower than expected from Raoult’s law. An example of negative deviation is a mixture of chloroform and acetone or a solution of water and hydrochloric acid.

Alternatively, positive deviation takes place when the cohesion between similar molecules is greater or that it exceeds adhesion between unlike or dissimilar molecules. Both components of the mixture can easily escape from the solution. An example of positive deviation includes the mixtures of benzene and methanol or ethanol and chloroform.

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