Ideal and Non-ideal solution Questions in English

(A) According to Raoult's law,the partial vapour pressure of a component is given by $P_i = P_i^{\circ} \times X_i$.
First,calculate the number of moles of benzene $(C_6H_6)$ and toluene $(C_7H_8)$:
$n_{\text{benzene}} = \frac{78\,g}{78\,g/mol} = 1\,mol$.
$n_{\text{toluene}} = \frac{46\,g}{92\,g/mol} = 0.5\,mol$.
Next,calculate the mole fraction of benzene $(X_B)$:
$X_B = \frac{n_{\text{benzene}}}{n_{\text{benzene}} + n_{\text{toluene}}} = \frac{1}{1 + 0.5} = \frac{1}{1.5} = \frac{2}{3}$.
Now,calculate the partial vapour pressure of benzene $(P_B)$:
$P_B = P_B^{\circ} \times X_B = 75\,torr \times \frac{2}{3} = 50\,torr$.

(B) Partial pressure of methanol $= 2.619 \, kPa$
Partial pressure of ethanol $= 4.556 \, kPa$
Total pressure $= \text{Partial pressure of methanol} + \text{Partial pressure of ethanol}$
Total pressure $= 2.619 + 4.556 = 7.175 \, kPa$
Let $y_{methanol}$ be the mole fraction of methanol in the vapour phase.
$y_{methanol} = \frac{P_{methanol}}{P_{total}} = \frac{2.619}{7.175} = 0.365$
Let $y_{ethanol}$ be the mole fraction of ethanol in the vapour phase.
$y_{ethanol} = \frac{P_{ethanol}}{P_{total}} = \frac{4.556}{7.175} = 0.635$
Thus,the composition of the vapour is $0.365$ methanol and $0.635$ ethanol.
Hence,the correct option is $B$.

(B) Raoult's law is valid for dilute solutions containing non-volatile,non-electrolyte solutes. In such solutions,the solute does not undergo association or dissociation,allowing for the determination of the correct molecular mass. Therefore,it is applicable to a non-electrolyte in a dilute solution.

(B) The correct answer is $(b)$.
In a mixture of benzene and methanol,methanol molecules are initially held together by strong hydrogen bonds.
When benzene is added,it disrupts these hydrogen bonds,resulting in weaker intermolecular forces in the mixture compared to the pure components.
This reduction in intermolecular attraction leads to a higher vapor pressure than predicted by Raoult's law,which is defined as a positive deviation.

(D) An ideal solution follows Raoult's law over the entire range of concentration and shows no change in enthalpy or volume upon mixing.
Benzene + toluene,$n$-hexane + $n$-heptane,and ethyl bromide + ethyl iodide are examples of ideal solutions because the intermolecular forces between the components are similar.
$CCl_4 + CHCl_3$ is a non-ideal solution because the intermolecular forces between $CCl_4$ and $CHCl_3$ are different from the forces in the pure components,leading to deviations from Raoult's law.
Therefore,the correct option is $D$.

(B) Chloroform $(CHCl_3)$ and acetone $(CH_3COCH_3)$ form a non-ideal solution showing negative deviation from Raoult's Law.
In this mixture,the intermolecular $A-B$ interactions (between chloroform and acetone) are stronger than the $A-A$ (chloroform-chloroform) and $B-B$ (acetone-acetone) interactions due to the formation of hydrogen bonding between the oxygen of acetone and the hydrogen of chloroform.
Due to these stronger interactions,the molecules come closer to each other,resulting in a decrease in the total volume of the mixture.
Therefore,$\Delta V_{mix} < 0$,which means the total volume of the solution will be less than the sum of the individual volumes $(30 \ mL + 50 \ mL = 80 \ mL)$.
Thus,the volume of the mixture will be $< 80 \ mL$.

(B) An ideal solution is formed by components that have similar molecular structures and polarities,resulting in similar intermolecular forces ($A-A$,$B-B$,and $A-B$ interactions are nearly equal).
$CCl_4 + SiCl_4$,$C_2H_5Br + C_2H_5I$,and $C_6H_{14} + C_7H_{16}$ are pairs of structurally similar compounds that form ideal solutions.
$H_2O$ and $C_4H_9OH$ (butanol) do not form an ideal solution because they exhibit strong intermolecular hydrogen bonding,leading to significant deviations from Raoult's Law.

(D) An ideal solution is defined as a solution that obeys Raoult's law over the entire range of concentration at all temperatures.
For an ideal solution,the enthalpy of mixing $(\Delta H_{mix})$ is $0$ and the volume of mixing $(\Delta V_{mix})$ is $0$.

(D) An ideal solution is formed by components that have similar molecular structures and polarities,resulting in similar intermolecular forces.
$C_2H_5Br$ and $C_2H_5I$,$C_6H_5Cl$ and $C_6H_5Br$,and $C_6H_6$ and $C_6H_5CH_3$ form ideal solutions because their components are structurally similar.
However,$C_2H_5I$ and $C_2H_5OH$ do not form an ideal solution because $C_2H_5OH$ exhibits strong hydrogen bonding,whereas $C_2H_5I$ does not.
This difference in intermolecular forces leads to significant deviations from Raoult's Law.

(D) Ideal solutions exhibit the following characteristics:
$1$. They follow Raoult's Law over the entire range of concentration.
$2$. The enthalpy of mixing is zero,i.e.,$\Delta_{mix} H = 0$. This means no heat is absorbed or evolved during the formation of the solution.
$3$. The volume change upon mixing is zero,i.e.,$\Delta_{mix} V = 0$. This implies that the total volume of the solution is equal to the sum of the volumes of the individual components.
$4$. The intermolecular forces of attraction between solute-solute and solvent-solvent are nearly equal to those between solute-solvent.

(C) The boiling point of a solution is inversely proportional to its vapour pressure.
When the boiling point of the mixture is greater than the boiling points of both individual components,it indicates that the vapour pressure of the solution is lower than that predicted by Raoult's Law.
This decrease in vapour pressure is characteristic of a non-ideal solution showing negative deviation from Raoult's Law.

(D) For a solution to show negative deviation from Raoult's Law:
$1$. The intermolecular forces of attraction between $A$ and $B$ are stronger than those between $A-A$ and $B-B$ ($A-B > A-A$ and $B-B$).
$2$. The enthalpy change of mixing is negative,$\Delta H_{mix} < 0$.
$3$. The volume change of mixing is negative,$\Delta V_{mix} < 0$.
Therefore,the correct condition is that the $A-B$ interaction is stronger than the $A-A$ and $B-B$ interactions.

(C) solution that obeys Raoult's law is known as an ideal solution.
Raoult's law states that for an ideal solution,the partial vapor pressure of each component is directly proportional to its mole fraction. Since this law is the defining characteristic of ideal solutions,the correct answer is $C$.

(A) An ideal solution is one that obeys Raoult's law over the entire range of concentration and shows no change in enthalpy or volume upon mixing.
Mixtures of $n$-hexane and $n$-heptane are very similar in molecular structure and polarity,resulting in intermolecular forces that are nearly identical to those in the pure components.
Therefore,the mixture of $n$-heptane and $n$-hexane forms a near-ideal solution.

(A) positive deviation from Raoult's law occurs when the solute-solvent intermolecular interactions are weaker than the solute-solute and solvent-solvent interactions.
In the mixture of acetone $(CH_3)_2CO$ and ethanol $(C_2H_5OH)$,the hydrogen bonding present in pure ethanol is disrupted by the addition of acetone,leading to weaker interactions in the mixture.
Therefore,this mixture exhibits a positive deviation from Raoult's law.

(D) An ideal solution is formed by components that are chemically similar in structure and polarity,leading to similar intermolecular forces of attraction.
$C_2H_5Br$ and $C_2H_5I$,$C_2H_5Cl$ and $C_2H_5Br$,and $C_6H_6$ and $C_6H_5CH_3$ are pairs of homologous series or structurally similar compounds,which form nearly ideal solutions.
However,$C_2H_5I$ (an alkyl halide) and $C_2H_5OH$ (an alcohol) have different polarities and intermolecular forces (alcohol exhibits hydrogen bonding),so they form a non-ideal solution.

(B) The enthalpy of solution formation is the sum of the enthalpy changes of the individual steps:
$\Delta H_{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3$
For an ideal solution,the net enthalpy change of mixing is zero,meaning $\Delta H_{mix} = 0$.
However,the question asks for the expression of $\Delta H_{soln}$ based on the given steps.
By Hess's Law,the total enthalpy change is the sum of the enthalpy changes of the intermediate steps:
$\Delta H_{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3$

(B) When acetone $(CH_3COCH_3)$ is mixed with chloroform $(CHCl_3)$,a strong intermolecular hydrogen bond is formed between the oxygen atom of acetone and the hydrogen atom of chloroform.
This interaction makes the attraction between the molecules of $A$ and $B$ stronger than the individual $A-A$ or $B-B$ interactions.
As a result,the escaping tendency of both components decreases,leading to a lower vapour pressure than expected for an ideal solution.
Therefore,this mixture shows a negative deviation from Raoult's law.

(A) An ideal solution is formed when the components obey Raoult's Law over the entire range of concentration.
For an ideal solution,the enthalpy of mixing $(\Delta H_{mix})$ is $0$ because the intermolecular forces between the components are similar to those in the pure components.
The volume of mixing $(\Delta V_{mix})$ is also $0$ because there is no change in the total volume upon mixing.
Therefore,$\Delta H_{mix} = \Delta V_{mix} = 0$.

(D) Benzene and toluene have similar molecular structures and polarities.
When mixed,the intermolecular forces of attraction between benzene-benzene,toluene-toluene,and benzene-toluene are nearly identical.
Therefore,the mixture forms an ideal solution and shows practically no deviation from Raoult's law.

(C) Solutions that do not obey Raoult's law over the entire range of concentration are known as Non-Ideal Solutions.
These solutions exhibit either positive or negative deviations from Raoult's law depending on the intermolecular forces between the solute and solvent molecules compared to those in the pure components.

(A) Positive deviation from Raoult's law occurs when the solute-solvent interactions are weaker than the solute-solute and solvent-solvent interactions.
Benzene-Chloroform mixture shows negative deviation from Raoult's law because of the formation of strong intermolecular hydrogen bonding or dipole-dipole interactions between the components,which leads to a decrease in vapor pressure.
Therefore,the mixture that does not show positive deviation is Benzene-Chloroform.

(D) . $CHCl_3 + (CH_3)_2CO$: The formation of strong intermolecular $H$-bonding between chloroform and acetone leads to negative deviation from Raoult's law.
$B$. $C_6H_6 + C_6H_5CH_3$: These are structurally similar and form an ideal solution.
$C$. $H_2O + HCl$: This mixture shows negative deviation due to strong solute-solvent interactions.
$D$. $CCl_4 + CHCl_3$: The interaction between $CCl_4$ and $CHCl_3$ is weaker than the pure component interactions,leading to positive deviation from Raoult's law.

(D) Ideal solutions exhibit the following characteristics:
$1$. They obey Raoult's Law,meaning the partial pressure of components $A$ and $B$ follows $P_A = P_A^o \times X_A$ and $P_B = P_B^o \times X_B$.
$2$. The enthalpy of mixing is zero,i.e.,$\Delta H_{mix} = 0$. This indicates that no heat is evolved or absorbed during the formation of the solution.
$3$. The volume change upon mixing is zero,i.e.,$\Delta V_{mix} = 0$. This means the total volume of the solution equals the sum of the volumes of the individual components.
$4$. The intermolecular forces of attraction between solute-solute and solvent-solvent are similar to those between solute-solvent.
Since options $A$,$B$,and $C$ are all properties of an ideal solution,the correct choice is $D$.

(A) For an ideal solution,the following conditions must be met:
$1$. It obeys Raoult's law over the entire range of concentration.
$2$. The enthalpy of mixing,$\Delta H_{mix} = 0$.
$3$. The volume of mixing,$\Delta V_{mix} = 0$.
$4$. The entropy of mixing,$\Delta S_{mix} > 0$ (since mixing is a spontaneous process).
Therefore,the statement $\Delta S_{mix} = 0$ is incorrect.

(A) The mixing of benzene and toluene does not involve any significant change in the intermolecular forces of attraction between the molecules.
Since the solute-solute,solvent-solvent,and solute-solvent interactions are nearly identical,they form an ideal solution.

(C) An ideal solution is one that obeys Raoult's law over the entire range of concentration and shows no change in enthalpy or volume upon mixing.
$1$. Water $+$ ethanol shows positive deviation from Raoult's law due to the disruption of hydrogen bonding.
$2$. Chloroform $+$ carbon tetrachloride shows positive deviation from Raoult's law.
$3$. Water $+$ hydrochloric acid shows negative deviation from Raoult's law.
$4$. Benzene $+$ toluene forms an ideal solution because the intermolecular forces of attraction between benzene-benzene,toluene-toluene,and benzene-toluene are nearly identical,as they are structurally similar (differing only by a $-CH_3$ group).

(A) When ethanol is mixed with cyclohexane,the strong hydrogen bonding between ethanol molecules is disrupted by the non-polar cyclohexane molecules.
As a result,the intermolecular forces between ethanol and cyclohexane $(A-B)$ are weaker than the forces between pure ethanol molecules $(A-A)$ or pure cyclohexane molecules $(B-B)$.
This leads to an increase in vapor pressure and a positive deviation from Raoult's law.
Therefore,the correct option is $A$.

(A) For an ideal solution,the following conditions must be satisfied:-
$1. \Delta H_{mix} = 0$ (Enthalpy of mixing is zero)
$2. \Delta V_{mix} = 0$ (Volume of mixing is zero)
$3. \Delta S_{mix} > 0$ (Entropy of mixing is positive)
$4. \Delta G_{mix} < 0$ (Gibbs free energy of mixing is negative)
Since $\Delta H_{mix} = 0$ is a defining characteristic of an ideal solution,option $A$ is correct.

(C) According to Raoult's Law for an ideal solution,the total vapor pressure $P_{total}$ is given by:
$P_{total} = P_A^o \chi_A + P_B^o \chi_B$
Given:
$P_A^o = 70 \ torr$
$\chi_B = 0.2$
$\chi_A = 1 - \chi_B = 1 - 0.2 = 0.8$
$P_{total} = 84 \ torr$
Substituting the values:
$84 = (70 \times 0.8) + (P_B^o \times 0.2)$
$84 = 56 + 0.2 P_B^o$
$84 - 56 = 0.2 P_B^o$
$28 = 0.2 P_B^o$
$P_B^o = \frac{28}{0.2} = 140 \ torr$
Thus,the vapor pressure of pure liquid $B$ is $140 \ torr$.

(B) $1$. Calculate the moles of each component:
$n_{\text{heptane}} = \frac{25 \ g}{100 \ g \ mol^{-1}} = 0.25 \ mol$
$n_{\text{octane}} = \frac{35 \ g}{114 \ g \ mol^{-1}} \approx 0.307 \ mol$
$2$. Calculate the mole fractions:
$n_{\text{total}} = 0.25 + 0.307 = 0.557 \ mol$
$x_{\text{heptane}} = \frac{0.25}{0.557} \approx 0.449$
$x_{\text{octane}} = \frac{0.307}{0.557} \approx 0.551$
$3$. Apply Raoult's Law:
$P_{\text{total}} = P^{\circ}_{\text{heptane}} \cdot x_{\text{heptane}} + P^{\circ}_{\text{octane}} \cdot x_{\text{octane}}$
$P_{\text{total}} = (105 \times 0.449) + (45 \times 0.551)$
$P_{\text{total}} = 47.145 + 24.795 = 71.94 \ kPa \approx 72.0 \ kPa$.

(D) Total moles $= 1 + 4 = 5 \ mol$.
Mole fraction of heptane $(X_A) = 1/5 = 0.2$.
Mole fraction of octane $(X_B) = 4/5 = 0.8$.
According to Raoult's Law,the total vapor pressure of the solution is given by:
$P_{total} = X_A P_A^0 + X_B P_B^0$
$P_{total} = (0.2 \times 92) + (0.8 \times 31)$
$P_{total} = 18.4 + 24.8 = 43.2 \ mm \ Hg$.

(B) For an ideal solution,the enthalpy of mixing is zero,i.e.,$\Delta H_{mix} = 0$.
The total enthalpy change of the solution is given by $\Delta H_{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3$.
Since $\Delta H_{soln} = 0$ for an ideal solution,we have $\Delta H_1 + \Delta H_2 + \Delta H_3 = 0$.
However,based on the standard thermodynamic cycle for solution formation,the enthalpy of solution is $\Delta H_{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3$.
If we define $\Delta H_3$ as the enthalpy of interaction between solute and solvent,then for an ideal solution,$\Delta H_1 + \Delta H_2 + \Delta H_3 = 0$.

(B) Positive deviation from Raoult's law occurs when the solute-solvent interactions are weaker than the solute-solute and solvent-solvent interactions. $A-B$ interactions are weaker than $A-A$ and $B-B$ interactions. In the case of $Benzene-Methanol$,the hydrogen bonding between methanol molecules is disrupted by the addition of benzene,leading to weaker interactions and positive deviation. Options $A$ and $C$ show negative deviation due to strong hydrogen bonding or ionization,while $D$ shows negative deviation due to hydrogen bonding between acetone and chloroform.

(B) Negative deviation from Raoult's law occurs when the solute-solvent interaction is stronger than the solute-solute and solvent-solvent interactions.
Acetone-Chloroform,Chloroform-Ether,and Chloroform-Benzene exhibit negative deviation due to hydrogen bonding or strong dipole-dipole interactions.
Acetone-Benzene shows positive deviation from Raoult's law because the interaction between acetone and benzene is weaker than the pure components' interactions.

(B) In the vapor phase,the mole fraction of a component is given by the ratio of its partial pressure to the total pressure of the solution.
Total pressure $P_{total} = P_{MeOH} + P_{EtOH} = 2.619 \ kPa + 4.556 \ kPa = 7.175 \ kPa$.
Mole fraction of $MeOH$ in vapor phase $(y_{MeOH}) = \frac{P_{MeOH}}{P_{total}} = \frac{2.619}{7.175} = 0.365$.
Mole fraction of $EtOH$ in vapor phase $(y_{EtOH}) = \frac{P_{EtOH}}{P_{total}} = \frac{4.556}{7.175} = 0.635$.

(C) Let $P_A^0$ and $P_B^0$ be the vapor pressures of pure $x$ and $y$ respectively.
For the first solution: $n_x = 1, n_y = 3$. Total moles = $4$.
Mole fractions: $X_x = 1/4, X_y = 3/4$.
$P_{total} = P_A^0 X_x + P_B^0 X_y \Rightarrow 550 = P_A^0(1/4) + P_B^0(3/4)$.
$P_A^0 + 3P_B^0 = 2200$ ... $(1)$
For the second solution: $n_x = 1, n_y = 4$. Total moles = $5$.
$P_{total} = 550 + 10 = 560 \ mm \ Hg$.
Mole fractions: $X_x = 1/5, X_y = 4/5$.
$560 = P_A^0(1/5) + P_B^0(4/5)$.
$P_A^0 + 4P_B^0 = 2800$ ... $(2)$
Subtracting $(1)$ from $(2)$:
$(P_A^0 + 4P_B^0) - (P_A^0 + 3P_B^0) = 2800 - 2200$.
$P_B^0 = 600 \ mm \ Hg$.
Substituting $P_B^0$ in $(1)$:
$P_A^0 + 3(600) = 2200 \Rightarrow P_A^0 = 2200 - 1800 = 400 \ mm \ Hg$.
Thus,the vapor pressures are $400 \ mm \ Hg$ and $600 \ mm \ Hg$.

(B) $1$. Calculate the moles of benzene $(C_6H_6)$: $n_{benzene} = \frac{78 \ g}{78 \ g/mol} = 1 \ mol$.
$2$. Calculate the moles of toluene $(C_7H_8)$: $n_{toluene} = \frac{46 \ g}{92 \ g/mol} = 0.5 \ mol$.
$3$. Calculate the mole fraction of benzene $(X_{benzene})$: $X_{benzene} = \frac{n_{benzene}}{n_{benzene} + n_{toluene}} = \frac{1}{1 + 0.5} = \frac{1}{1.5} = \frac{2}{3}$.
$4$. Calculate the partial vapor pressure of benzene $(P_{benzene})$ using Raoult's Law: $P_{benzene} = P^0_{benzene} \times X_{benzene} = 75 \ torr \times \frac{2}{3} = 50 \ torr$.

(A) According to Raoult's law,the total pressure $P$ of a solution containing two volatile components $A$ and $B$ is given by:
$P = P_A + P_B$
Since $P_A = X_A P_A^o$ and $P_B = X_B P_B^o$,we have:
$P = X_A P_A^o + X_B P_B^o$
Substituting $X_B = 1 - X_A$:
$P = X_A P_A^o + (1 - X_A) P_B^o$
$P = X_A P_A^o + P_B^o - X_A P_B^o$
$P = X_A (P_A^o - P_B^o) + P_B^o$
This equation is derived directly from Raoult's law,which defines an $Ideal$ solution.

(D) When the mixture of benzene and toluene is boiled at $88 \, ^\circ C$,the total vapor pressure of the solution becomes equal to the external pressure,which is $1 \, \text{atm} = 760 \, \text{torr}$.
According to Raoult's Law for an ideal solution: $P_S = P_A^0 X_A + P_B^0 X_B$
Here,$P_S = 760 \, \text{torr}$,$P_A^0 = 900 \, \text{torr}$ (benzene),$P_B^0 = 360 \, \text{torr}$ (toluene),and $X_B = 1 - X_A$.
Substituting the values: $760 = 900 X_A + 360(1 - X_A)$
$760 = 900 X_A + 360 - 360 X_A$
$760 - 360 = 540 X_A$
$400 = 540 X_A$
$X_A = \frac{400}{540} \approx 0.7407 \approx 0.740$.

(A) non-ideal solution shows negative deviation from Raoult's law when the intermolecular forces of attraction between the solute $(A)$ and solvent $(B)$ are stronger than the forces of attraction between the pure components ($A-A$ and $B-B$).
This leads to a decrease in volume $(\Delta V_{mix} < 0)$ and a release of heat $(\Delta H_{mix} < 0)$.

(C) The boiling point is the temperature at which the total vapor pressure of the solution equals the external pressure.
Given: $P_{\text{total}} = 1 \, \text{atm} = 760 \, \text{torr}$.
Let $x_B$ be the mole fraction of benzene $(C_6H_6)$ and $x_T$ be the mole fraction of toluene $(C_7H_8)$.
Since $x_B + x_T = 1$,we have $x_T = 1 - x_B$.
According to Raoult's law: $P_{\text{total}} = P_B^0 x_B + P_T^0 x_T$.
Substituting the values: $760 = 900 x_B + 360 (1 - x_B)$.
$760 = 900 x_B + 360 - 360 x_B$.
$760 - 360 = 540 x_B$.
$400 = 540 x_B$.
$x_B = \frac{400}{540} \approx 0.74$.

(D) Positive deviation from Raoult's law occurs when the solute-solvent intermolecular forces are weaker than the solute-solute and solvent-solvent forces.
$A$,$B$,and $C$ (Benzene-Chloroform,Benzene-Acetone,and Benzene-Ethanol) exhibit positive deviations because the interaction between the components is weaker than the pure components.
$D$ (Benzene-Carbon tetrachloride) forms an ideal solution because the intermolecular forces between $C_6H_6$ and $CCl_4$ are similar to those in the pure components.
Therefore,it does not show positive deviation.

(C) An ideal solution is defined as a solution that obeys Raoult's law over the entire range of concentration. For an ideal solution,the following conditions must be met:
$1$. The enthalpy of mixing,$\Delta H_{mix} = 0$.
$2$. The volume of mixing,$\Delta V_{mix} = 0$.
$3$. The entropy of mixing,$\Delta S_{mix} > 0$.
$4$. The Gibbs energy of mixing,$\Delta G_{mix} < 0$.
Since $\Delta G_{mix} = \Delta H_{mix} - T \Delta S_{mix}$,and $\Delta H_{mix} = 0$ while $\Delta S_{mix} > 0$,it follows that $\Delta G_{mix} < 0$.
Therefore,the correct statement is that the enthalpy of mixing is zero.

(B) According to Raoult's Law,$P_{total} = P_A^0 X_A + P_B^0 X_B$.
For the $1:1$ molar ratio,$X_A = 1/2$ and $X_B = 1/2$:
$400 = P_A^0(1/2) + P_B^0(1/2) \implies P_A^0 + P_B^0 = 800$ (Equation $1$).
For the $1:2$ molar ratio,$X_A = 1/3$ and $X_B = 2/3$:
$350 = P_A^0(1/3) + P_B^0(2/3) \implies P_A^0 + 2P_B^0 = 1050$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$:
$(P_A^0 + 2P_B^0) - (P_A^0 + P_B^0) = 1050 - 800 \implies P_B^0 = 250 \ mm \ Hg$.
Substituting $P_B^0 = 250$ into Equation $1$:
$P_A^0 + 250 = 800 \implies P_A^0 = 550 \ mm \ Hg$.
Thus,the vapor pressures are $550 \ mm \ Hg$ and $250 \ mm \ Hg$.

(D) An ideal binary liquid solution obeys Raoult's Law,which states that $P_A = P_A^0 X_A$ and $P_B = P_B^0 X_B$.
Since $P_{total} = P_A + P_B = P_A^0 X_A + P_B^0 (1 - X_A) = (P_A^0 - P_B^0) X_A + P_B^0$,the total pressure is a linear function of the mole fraction.
Therefore,graphs of $P_A$ vs $X_A$,$P_B$ vs $X_B$,and $P_{total}$ vs $X_A$ are all linear for an ideal solution.
Option $D$ states that the graph of $P_{total}$ vs $X_A$ is non-linear,which contradicts the behavior of an ideal solution.

(D) non-ideal solution shows negative deviation when the solute-solvent interaction is stronger than the solute-solute and solvent-solvent interactions.
In the mixture of $CH_3COCH_3$ (acetone) and $CHCl_3$ (chloroform),a strong hydrogen bond is formed between the oxygen atom of acetone and the hydrogen atom of chloroform.
This leads to a decrease in volume and enthalpy of mixing,resulting in negative deviation from Raoult's Law.
Therefore,the correct option is $(D)$.

(D) An ideal solution is defined by the following characteristics:
$1$. $\Delta H_{mix} = 0$ (Enthalpy of mixing is zero).
$2$. $\Delta V_{mix} = 0$ (Volume of mixing is zero).
$3$. $\Delta U_{mix} = 0$ (Internal energy of mixing is zero).
$4$. The solution obeys Raoult's law over the entire range of concentration,so $\Delta P = 0$.
However,for any spontaneous mixing process,the entropy of mixing $\Delta S_{mix}$ is always positive.
Since $\Delta G_{mix} = \Delta H_{mix} - T\Delta S_{mix}$,and $\Delta H_{mix} = 0$ while $\Delta S_{mix} > 0$,it follows that $\Delta G_{mix} = -T\Delta S_{mix} < 0$.
Therefore,the statement $\Delta G_{mix} = 0$ is incorrect.

(C) According to Raoult's Law,the partial pressure of a component in the vapour phase is given by $P_i = P^o_i X_i$.
Since the mixture is $1:1$ molar,the mole fractions in the liquid phase are equal $(X_{\text{benzene}} = X_{\text{toluene}} = 0.5)$.
The total pressure $P = P^o_{\text{benzene}} X_{\text{benzene}} + P^o_{\text{toluene}} X_{\text{toluene}}$.
The mole fraction in the vapour phase is $Y_i = \frac{P_i}{P} = \frac{P^o_i X_i}{P}$.
Since $X_{\text{benzene}} = X_{\text{toluene}}$,the component with the higher pure vapour pressure $(P^o)$ will have a higher mole fraction in the vapour phase.
Given $P^o_{\text{benzene}} = 12.8\ kPa$ and $P^o_{\text{toluene}} = 3.85\ kPa$,benzene is more volatile.
Therefore,the vapour will contain a higher percentage of benzene.

(D) For an ideal solution,the enthalpy change of mixing $(\Delta H_{mix})$ is $0$,the volume change of mixing $(\Delta V_{mix})$ is $0$,and the pressure deviation $(\Delta P)$ is $0$.
However,for any spontaneous mixing process,the entropy change of mixing $(\Delta S_{mix})$ is always greater than $0$ $(\Delta S_{mix} > 0)$.