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) ${\Delta _{{\text{mix}}}}{\text{H = 0,}}{\Delta _{{\text{mix}}}}{\text{V = 0,}}{{\text{P}}_{{\text{Total}}}} = {\text{p}}_{\text{B}}^ \circ {{\text{x}}_{\text{B}}}$

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.