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Hint: For our knowledge, Raoult’s is the first chemist to propose the relation between mole fractions and vapour pressure of the two components in the solution. This law is popularly known as Raoult’s law.
Raoult’s law states that in the case of a solution of volatile liquids, the partial vapour pressure of each compound of the solution is directly proportional to its mole fraction.
For example, $A$ and $B$ are the two compounds in solution.
The partial vapour pressure of component $A$
${P_A}\;\propto \;{x_A}$
${P_A}\; = k\;{x_A}$
$k\; = \;P_A^ \circ $, when ${x_A}\;{\text{ = 1}}$
${P_A}\; = \;P_A^ \circ \;{x_A}$
For component B,
${P_B}\; = \;P_B^ \circ \;{x_B}$
Here, ${P_A}$ is the partial vapour pressure of the compound $A$
${P_B}$ is the partial vapour pressure of the compound $B$
$P_A^ \circ $ is the vapour pressure of the pure compound $A$
$P_B^ \circ $ is the vapour pressure of the pure compound $B$
${x_A}$ is the mole fraction of the compound $A$
${x_B}$ is the mole fraction of the compound $B$
Complete step by step answer:
We should remember that, in Raoult’s law, two deviations are arising due to the vapour pressure of an ideal solution. The two deviations are positive and negative deviations.
The total vapour pressure of the solution is greater than the sum of the vapour pressure in case of an ideal solution. This kind of deviation occurring in Raoult’s law is known as a positive deviation.
\[P\; = \;{P_A} + {P_B} > P_A^ \circ \;{x_A}\;{\text{ + }}P_B^ \circ \;{x_B}\]
The total vapour pressure of the solution is lesser than the sum of the vapour pressure in the case of an ideal solution. This kind of deviation occurring in Raoult’s law is known as a negative deviation.
\[P\; = \;{P_A} + {P_B} < P_A^ \circ \;{x_A}\;{\text{ + }}P_B^ \circ \;{x_B}\]
In Raoult’s law for positive deviation, ${\Delta _{mix}}H$ value became a positive sign.
In Raoult’s law for negative deviation, ${\Delta _{mix}}H$ value became a negative sign.
Note: According to Dalton’s law, the total pressure of the solution is equal to the sum of the partial pressure of a component in the solution. In non-ideal solutions only positive and negative deviations from Raoult’s law. A positive deviation in Raoult’s law is due to the hydrogen bonding in-between the molecules. These hydrogen bonds increase the attraction between the molecules in solution. Hence, vapour pressure of the solution is greater than the calculation. In negative deviation, the attraction in-between the molecule is reduced. Hence, the vapour pressure of the solution is lesser than the calculation of Raoult’s law.
Raoult’s law states that in the case of a solution of volatile liquids, the partial vapour pressure of each compound of the solution is directly proportional to its mole fraction.
For example, $A$ and $B$ are the two compounds in solution.
The partial vapour pressure of component $A$
${P_A}\;\propto \;{x_A}$
${P_A}\; = k\;{x_A}$
$k\; = \;P_A^ \circ $, when ${x_A}\;{\text{ = 1}}$
${P_A}\; = \;P_A^ \circ \;{x_A}$
For component B,
${P_B}\; = \;P_B^ \circ \;{x_B}$
Here, ${P_A}$ is the partial vapour pressure of the compound $A$
${P_B}$ is the partial vapour pressure of the compound $B$
$P_A^ \circ $ is the vapour pressure of the pure compound $A$
$P_B^ \circ $ is the vapour pressure of the pure compound $B$
${x_A}$ is the mole fraction of the compound $A$
${x_B}$ is the mole fraction of the compound $B$
Complete step by step answer:
We should remember that, in Raoult’s law, two deviations are arising due to the vapour pressure of an ideal solution. The two deviations are positive and negative deviations.
The total vapour pressure of the solution is greater than the sum of the vapour pressure in case of an ideal solution. This kind of deviation occurring in Raoult’s law is known as a positive deviation.
\[P\; = \;{P_A} + {P_B} > P_A^ \circ \;{x_A}\;{\text{ + }}P_B^ \circ \;{x_B}\]
The total vapour pressure of the solution is lesser than the sum of the vapour pressure in the case of an ideal solution. This kind of deviation occurring in Raoult’s law is known as a negative deviation.
\[P\; = \;{P_A} + {P_B} < P_A^ \circ \;{x_A}\;{\text{ + }}P_B^ \circ \;{x_B}\]
In Raoult’s law for positive deviation, ${\Delta _{mix}}H$ value became a positive sign.
In Raoult’s law for negative deviation, ${\Delta _{mix}}H$ value became a negative sign.
Note: According to Dalton’s law, the total pressure of the solution is equal to the sum of the partial pressure of a component in the solution. In non-ideal solutions only positive and negative deviations from Raoult’s law. A positive deviation in Raoult’s law is due to the hydrogen bonding in-between the molecules. These hydrogen bonds increase the attraction between the molecules in solution. Hence, vapour pressure of the solution is greater than the calculation. In negative deviation, the attraction in-between the molecule is reduced. Hence, the vapour pressure of the solution is lesser than the calculation of Raoult’s law.
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What is meant by positive and negative deviation from Raoult’s law and how is the sign of ${\Delta _{mix}}H$ related to positive and negative deviation from Raoult’s law?
SOLUTIONS Chemistry Class 12 - NCERT EXERCISE 1.14 | Class 12 Chemistry Chapter 1 | Nandini Ma'am
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