
What are the conditions for an ideal solution which obeys Raoult’s law over the entire range of concentration?
(A) ${\Delta _{{\text{mix}}}}{\text{H = 0,}}{\Delta _{{\text{mix}}}}{\text{V = 0,}}{{\text{P}}_{{\text{Total}}}} = {\text{p}}_{\text{A}}^ \circ {{\text{x}}_{\text{A}}}{\text{ + p}}_{\text{B}}^ \circ {{\text{x}}_{\text{B}}}$
(B) ${\Delta _{{\text{mix}}}}{\text{H = + ve,}}{\Delta _{{\text{mix}}}}{\text{V = 0,}}{{\text{P}}_{{\text{Total}}}} = {\text{p}}_{\text{A}}^ \circ {{\text{x}}_{\text{A}}}{\text{ + p}}_{\text{B}}^ \circ {{\text{x}}_{\text{B}}}$
(C) ${\Delta _{{\text{mix}}}}{\text{H = 0,}}{\Delta _{{\text{mix}}}}{\text{V = + ve,}}{{\text{P}}_{{\text{Total}}}} = {\text{p}}_{\text{A}}^ \circ {{\text{x}}_{\text{A}}}{\text{ + p}}_{\text{B}}^ \circ {{\text{x}}_{\text{B}}}$
(D) ${\Delta _{{\text{mix}}}}{\text{H = 0,}}{\Delta _{{\text{mix}}}}{\text{V = 0,}}{{\text{P}}_{{\text{Total}}}} = {\text{p}}_{\text{B}}^ \circ {{\text{x}}_{\text{B}}}$
Answer
504k+ views
Hint: In physical chemistry, Raoult’s law states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.
Complete answer:
In an ideal solution, the solvent-solute interaction is similar to the solvent-solvent or solute-solute interaction. This means both the solute and solvent take the same amount of energy to escape to the vapor phase as when they are in their states.
For an ideal solution which obeys Raoult’s law the following are valid: ${\Delta _{{\text{mix}}}}{\text{V = 0}}$
Heat is neither released nor absorbed during the reaction i.e. ${\Delta _{{\text{mix}}}}{\text{H = 0}}$
The volume of the solution will remain the same i.e.
For an ideal solution, $\Delta {\text{H}}$ and $\Delta {\text{V}}$ for mixing should be zero. ${{\text{P}}_{{\text{Total}}}} = {{\text{p}}_{\text{A}}}{\text{ + }}{{\text{p}}_{\text{B}}}$ and the ${\text{A - A, B - B}}$ and ${\text{A - B}}$ interactions should be nearly the same.
Mathematically, Raoult’s law is expressed as:
${{\text{P}}_{{\text{solution}}}} = {{\text{x}}_{{\text{solvent}}}}{\text{p}}_{{\text{solvent}}}^ \circ $
Where,
${{\text{P}}_{{\text{solution}}}} = $ vapor pressure of the solution
${{\text{x}}_{{\text{solvent}}}} = $ mole fraction of the solvent
${\text{p}}_{{\text{solvent}}}^ \circ = $ vapor pressure of the pure solvent
For an ideal solution which obeys Raoult’s law over the entire range of concentration, ${\Delta _{{\text{mix}}}}{\text{H = 0,}}{\Delta _{{\text{mix}}}}{\text{V = 0,}}{{\text{P}}_{{\text{Total}}}} = {\text{p}}_{\text{A}}^ \circ {{\text{x}}_{\text{A}}}{\text{ + p}}_{\text{B}}^ \circ {{\text{x}}_{\text{B}}}$
Therefore, Option A is the correct answer.
Note:
Raoult’s law assumes that the intermolecular forces that are present between different molecules and similar molecules are equal. This law is apt for describing ideal solutions. Many of the liquids that are in the mixture do not have the same uniformity in terms of attractive forces, so these types of solutions tend to deviate away from the law.
Complete answer:
In an ideal solution, the solvent-solute interaction is similar to the solvent-solvent or solute-solute interaction. This means both the solute and solvent take the same amount of energy to escape to the vapor phase as when they are in their states.
For an ideal solution which obeys Raoult’s law the following are valid: ${\Delta _{{\text{mix}}}}{\text{V = 0}}$
Heat is neither released nor absorbed during the reaction i.e. ${\Delta _{{\text{mix}}}}{\text{H = 0}}$
The volume of the solution will remain the same i.e.
For an ideal solution, $\Delta {\text{H}}$ and $\Delta {\text{V}}$ for mixing should be zero. ${{\text{P}}_{{\text{Total}}}} = {{\text{p}}_{\text{A}}}{\text{ + }}{{\text{p}}_{\text{B}}}$ and the ${\text{A - A, B - B}}$ and ${\text{A - B}}$ interactions should be nearly the same.
Mathematically, Raoult’s law is expressed as:
${{\text{P}}_{{\text{solution}}}} = {{\text{x}}_{{\text{solvent}}}}{\text{p}}_{{\text{solvent}}}^ \circ $
Where,
${{\text{P}}_{{\text{solution}}}} = $ vapor pressure of the solution
${{\text{x}}_{{\text{solvent}}}} = $ mole fraction of the solvent
${\text{p}}_{{\text{solvent}}}^ \circ = $ vapor pressure of the pure solvent
For an ideal solution which obeys Raoult’s law over the entire range of concentration, ${\Delta _{{\text{mix}}}}{\text{H = 0,}}{\Delta _{{\text{mix}}}}{\text{V = 0,}}{{\text{P}}_{{\text{Total}}}} = {\text{p}}_{\text{A}}^ \circ {{\text{x}}_{\text{A}}}{\text{ + p}}_{\text{B}}^ \circ {{\text{x}}_{\text{B}}}$
Therefore, Option A is the correct answer.
Note:
Raoult’s law assumes that the intermolecular forces that are present between different molecules and similar molecules are equal. This law is apt for describing ideal solutions. Many of the liquids that are in the mixture do not have the same uniformity in terms of attractive forces, so these types of solutions tend to deviate away from the law.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

