Ideal and Non-ideal solution Questions in English

(B) For an ideal solution,the following conditions must be met:
$1$. $\Delta _{mix} H = 0$ (no heat is absorbed or evolved).
$2$. $\Delta _{mix} V = 0$ (no change in volume upon mixing).
$3$. It must obey Raoult's Law over the entire range of concentration.
$4$. For any spontaneous mixing process,the entropy of mixing must be positive,i.e.,$\Delta _{mix} S > 0$.
Therefore,the condition $\Delta _{mix} S = 0$ is not satisfied by an ideal solution.

(D) According to Raoult's Law,the total pressure $P_T$ of an ideal binary solution is given by the sum of the partial pressures of the components:
$P_T = P_A + P_B$
We know that $P_A = p_A x_A$ and $P_B = p_B x_B$,where $p_A$ and $p_B$ are the vapour pressures of pure components.
Substituting these into the equation:
$P_T = p_A x_A + p_B x_B$
Since $x_A + x_B = 1$,we have $x_B = 1 - x_A$.
Substituting $x_B$ in the equation:
$P_T = p_A x_A + p_B(1 - x_A)$
$P_T = p_A x_A + p_B - p_B x_A$
$P_T = p_B + x_A(p_A - p_B)$

(C) solution of acetone in ethanol shows a positive deviation from Raoult's law.
In pure ethanol,molecules are held together by strong hydrogen bonding.
When acetone is added,it disrupts these hydrogen bonds,leading to weaker solute-solvent interactions compared to the original solvent-solvent interactions.
This results in an increase in vapor pressure,which is characteristic of a positive deviation from Raoult's law.

(B) In a mixture of $n$-heptane and ethanol,the intermolecular forces between $n$-heptane and ethanol molecules are weaker than the forces between $n$-heptane$-n$-heptane molecules and ethanol-ethanol molecules.
Due to the presence of hydrogen bonding in pure ethanol,the $n$-heptane molecules disrupt these interactions.
As a result,the total vapor pressure of the solution is higher than that predicted by Raoult's Law.
Therefore,the solution is non-ideal and shows a positive $(+ve)$ deviation from Raoult's Law.

(B) According to Raoult's law,$P_{\text{total}} = P_X^{\circ} X_X + P_Y^{\circ} X_Y.$
For the first case: $X_X = \frac{1}{1+3} = \frac{1}{4}$ and $X_Y = \frac{3}{4}.$
$550 = P_X^{\circ} \times \frac{1}{4} + P_Y^{\circ} \times \frac{3}{4} \implies P_X^{\circ} + 3P_Y^{\circ} = 2200 \dots (i)$
For the second case,total moles of $Y = 3+1 = 4 \ mol.$ Total moles of solution = $1+4 = 5 \ mol.$
$X_X = \frac{1}{5}$ and $X_Y = \frac{4}{5}.$
New vapour pressure = $550 + 10 = 560 \ mm \ Hg.$
$560 = P_X^{\circ} \times \frac{1}{5} + P_Y^{\circ} \times \frac{4}{5} \implies P_X^{\circ} + 4P_Y^{\circ} = 2800 \dots (ii)$
Subtracting equation $(i)$ from $(ii)$:
$(P_X^{\circ} + 4P_Y^{\circ}) - (P_X^{\circ} + 3P_Y^{\circ}) = 2800 - 2200$
$P_Y^{\circ} = 600 \ mm \ Hg.$
Substituting $P_Y^{\circ}$ in equation $(i)$:
$P_X^{\circ} + 3(600) = 2200$
$P_X^{\circ} = 2200 - 1800 = 400 \ mm \ Hg.$
Thus,the vapour pressures are $400 \ mm \ Hg$ and $600 \ mm \ Hg$ respectively.

(A) Step $1$: Calculate the number of moles of each component.
$n_{\text{heptane}} = \frac{25.0 \ g}{100 \ g \ mol^{-1}} = 0.25 \ mol$
$n_{\text{octane}} = \frac{35 \ g}{114 \ g \ mol^{-1}} \approx 0.307 \ mol$
Step $2$: Calculate the mole fractions.
$x_{\text{heptane}} = \frac{0.25}{0.25 + 0.307} = \frac{0.25}{0.557} \approx 0.4488$
$x_{\text{octane}} = \frac{0.307}{0.25 + 0.307} = \frac{0.307}{0.557} \approx 0.5512$
Step $3$: Apply Raoult's Law for the total vapour pressure.
$P_{\text{total}} = P_{\text{heptane}}^{\circ} x_{\text{heptane}} + P_{\text{octane}}^{\circ} x_{\text{octane}}$
$P_{\text{total}} = (105 \ kPa \times 0.4488) + (45 \ kPa \times 0.5512)$
$P_{\text{total}} = 47.124 + 24.804 = 71.928 \ kPa \approx 72.0 \ kPa$.

(B) For an ideal solution,the boiling point $(T_b)$ depends on the mole fraction $(x)$ of the components.
Let the mass of each component be $m$.
Moles of hexane $(n_1)$ = $m / 86$ and moles of heptane $(n_2)$ = $m / 100$.
Mole fraction of hexane $(x_1)$ = $n_1 / (n_1 + n_2) = (m/86) / (m/86 + m/100) = 100 / 186 \approx 0.537$.
Mole fraction of heptane $(x_2)$ = $1 - 0.537 = 0.463$.
The boiling point of the mixture is given by $T_b = x_1 T_{b1} + x_2 T_{b2}$ (approximately for ideal mixtures).
$T_b = (0.537 \times 69) + (0.463 \times 98) \approx 37.05 + 45.37 = 82.42 \ ^oC$.
Since the mole fraction of the lower boiling component (hexane) is higher in the equal mass mixture $(0.537)$ compared to the equimolar mixture $(0.5)$,the boiling point will be slightly higher than the equimolar boiling point of $80 \ ^oC$ but less than $98 \ ^oC$.
Thus,the boiling point is between $80 \ ^oC$ and $98 \ ^oC$.

(D) The mole fraction of $A$ $(x_A)$ is $10 / (10 + 10) = 0.5$ and $x_B = 0.5$.
According to Raoult's law,the expected vapour pressure is $P_{ideal} = P_A^0 x_A + P_B^0 x_B = (200 \times 0.5) + (100 \times 0.5) = 100 + 50 = 150 \ mm \ Hg$.
Since the observed vapour pressure $(160 \ mm \ Hg)$ is greater than the ideal vapour pressure $(150 \ mm \ Hg)$,the solution shows a positive deviation from Raoult's law.
For solutions showing positive deviation: $\Delta H_{mix} > 0$,$\Delta V_{mix} > 0$,and $\Delta G_{mix} < 0$.
Since $\Delta H_{mix} > 0$ (endothermic process),the heat is absorbed from the surroundings,so $\Delta S_{surrounding} = -\Delta H_{mix} / T < 0$.
None of the given options $(A, B, C)$ are correct.

(D) When two liquids are mixed,if the intermolecular forces of attraction between the components are weaker than the forces in the pure components,the mixture shows positive deviation from Raoult's law. In such cases,the volume of the solution increases $(V_{mix} > 0)$.
Conversely,if the forces are stronger (e.g.,due to hydrogen bonding),the mixture shows negative deviation,resulting in contraction in volume $(V_{mix} < 0)$.
$CHCl_3 + CH_3COCH_3$ forms hydrogen bonds between the chloroform hydrogen and the acetone oxygen,leading to negative deviation and contraction.
$H_2O + HCl$ and $H_2O + HNO_3$ show strong interactions due to ionization and hydrogen bonding,leading to contraction.
$H_2O + C_2H_5OH$ forms a mixture that exhibits positive deviation from Raoult's law because the hydrogen bonding between water and ethanol is weaker than the hydrogen bonding in pure water,leading to an increase in volume rather than contraction.

(A) Acetone and chloroform interact to form intermolecular hydrogen bonding between the oxygen atom of acetone and the hydrogen atom of chloroform.
This strong interaction leads to a decrease in the total volume of the solution compared to the sum of the individual volumes.
Therefore,the total volume of the solution is $< 50 \ mL$.

(A) According to Raoult's law,the partial pressure of a component in a solution is $P_A = P_A^{\circ} X_A$.
According to Dalton's law of partial pressures,the partial pressure of a component in the vapour phase is $P_A = P_T Y_A$.
Equating these,we get $P_A^{\circ} X_A = P_T Y_A$,which implies $\frac{Y_A}{X_A} = \frac{P_A^{\circ}}{P_T}$.
Since $A$ is more volatile than $B$,its pure vapour pressure $P_A^{\circ}$ is greater than the total pressure $P_T$ of the solution $(P_A^{\circ} > P_T)$.
Therefore,$\frac{Y_A}{X_A} > 1$,which means $Y_A > X_A$ or $X_A < Y_A$.

(C) According to Raoult's law and Dalton's law,the partial pressure of component $A$ is given by $P_A = X_A P_A^o$ and also $P_A = Y_A P_T$.
Equating these,we get $X_A P_A^o = Y_A P_T$.
Rearranging,we have $\frac{Y_A}{X_A} = \frac{P_A^o}{P_T}$.
Since $A$ is more volatile than $B$,$P_A^o > P_T$,which implies $\frac{P_A^o}{P_T} > 1$.
Therefore,$\frac{Y_A}{X_A} > 1$,which means $Y_A > X_A$ or $X_A < Y_A$.

(A) In pure ethanol,molecules are held together by strong intermolecular $H$-bonding.
When acetone is added to ethanol,the acetone molecules get in between the ethanol molecules and break some of the $H$-bonds between them.
This results in a weakening of the intermolecular forces of attraction in the mixture compared to the pure components.
Consequently,the vapor pressure of the solution increases,leading to a positive deviation from Raoult's law.

(C) The mixture of $CCl_4$ (carbon tetrachloride) and benzene $(C_6H_6)$ forms an ideal solution.
For an ideal solution,the enthalpy of mixing $(\Delta H_{mix})$ is $0$ and the volume of mixing $(\Delta V_{mix})$ is $0$.
Therefore,the total volume of the solution is equal to the sum of the volumes of the individual components.
Total volume = $10 \ mL 10 \ mL = 20 \ mL$.

(A) Aniline and phenol form a non-ideal solution showing negative deviation from Raoult's Law due to hydrogen bonding between the amine group of aniline and the hydroxyl group of phenol.
For a solution showing negative deviation,the total vapour pressure $(P_{total})$ is less than the vapour pressure calculated by Raoult's Law $(P_{ideal})$.
First,calculate the mole fractions:
$n_{aniline} = 10 \, moles$,$n_{phenol} = 20 \, moles$.
Total moles = $10 + 20 = 30 \, moles$.
$x_{aniline} = 10 / 30 = 1/3$,$x_{phenol} = 20 / 30 = 2/3$.
$P_{ideal} = x_{aniline} \times P^0_{aniline} + x_{phenol} \times P^0_{phenol} = (1/3 \times 90) + (2/3 \times 87) = 30 + 58 = 88 \, mmHg$.
Since the solution shows negative deviation,$P_{total} < 88 \, mmHg$.
Among the given options,only $80 \, mmHg$ is less than $88 \, mmHg$.

(A) According to Raoult's law,the ideal total vapour pressure $(P_{total, ideal})$ is given by:
$P_{total, ideal} = X_x P_x^o + X_y P_y^o$
Given: $n_x = 1 \, mol$,$n_y = 2 \, mol$.
Total moles = $1 + 2 = 3 \, mol$.
Mole fractions: $X_x = 1/3$,$X_y = 2/3$.
$P_{total, ideal} = (1/3 \times 150) + (2/3 \times 300) = 50 + 200 = 250 \, torr$.
The observed total vapour pressure is $240 \, torr$.
Since $P_{observed} < P_{total, ideal}$ $(240 \, torr < 250 \, torr)$,the mixture shows a negative deviation from Raoult's law.

(C) According to Raoult's law,the ideal vapour pressure $(P_{ideal})$ is calculated as:
$P_{ideal} = X_A P_A^0 + X_B P_B^0$
Given: $n_A = 1 \ mol$,$n_B = 2 \ mol$.
Total moles = $1 + 2 = 3 \ mol$.
Mole fractions: $X_A = 1/3$,$X_B = 2/3$.
$P_{ideal} = (1/3 \times 45) + (2/3 \times 36) = 15 + 24 = 39 \ torr$.
The observed vapour pressure is $42 \ torr$.
Since $P_{observed} > P_{ideal}$,the solution shows positive deviation from Raoult's law.
Solutions showing positive deviation exhibit an increase in volume upon mixing $(\Delta V_{mix} > 0)$ and may form a minimum boiling azeotrope.

(D) An ideal solution is defined as a solution that obeys Raoult's law over the entire range of concentration.
For an ideal solution,the enthalpy of mixing,$\Delta H_{mixing}$,is equal to $0$.
Furthermore,the volume of mixing,$\Delta V_{mixing}$,is also equal to $0$.
This occurs because the intermolecular forces of attraction between the solute-solute $(A-A)$,solvent-solvent $(B-B)$,and solute-solvent $(A-B)$ molecules are identical in nature and magnitude.

(C) Let the vapour pressure of pure $A$ be $P_A^o$ and pure $B$ be $P_B^o$.
In the first solution,the mole fraction of $A$ is $x_A = \frac{1}{1+2} = \frac{1}{3}$ and the mole fraction of $B$ is $x_B = \frac{2}{1+2} = \frac{2}{3}$.
According to Raoult's law,the total vapour pressure is $P_{total} = P_A^o x_A + P_B^o x_B$.
$250 = \frac{1}{3} P_A^o + \frac{2}{3} P_B^o \implies P_A^o + 2P_B^o = 750$ . . . $(i)$
In the second solution,$1\,mol$ of $A$ is added,so total moles of $A = 2$ and $B = 2$.
The mole fraction of $A$ is $x_A = \frac{2}{2+2} = \frac{1}{2}$ and the mole fraction of $B$ is $x_B = \frac{2}{2+2} = \frac{1}{2}$.
$300 = \frac{1}{2} P_A^o + \frac{1}{2} P_B^o \implies P_A^o + P_B^o = 600$ . . . $(ii)$
Subtracting equation $(ii)$ from equation $(i)$:
$(P_A^o + 2P_B^o) - (P_A^o + P_B^o) = 750 - 600$
$P_B^o = 150\,mm\,Hg$
Substituting $P_B^o = 150$ into equation $(ii)$:
$P_A^o + 150 = 600 \implies P_A^o = 450\,mm\,Hg$.
Thus,the vapour pressures of pure $A$ and $B$ are $450\,mm\,Hg$ and $150\,mm\,Hg$ respectively.

(B) Given: $\chi_A = 0.4$,$\chi_B = 0.6$,$P_A^o = 7 \times 10^3 \ Pa$,$P_B^o = 12 \times 10^3 \ Pa$.
Total pressure of the solution is $P_T = \chi_A P_A^o + \chi_B P_B^o$.
$P_T = (0.4 \times 7 \times 10^3) + (0.6 \times 12 \times 10^3) = 2.8 \times 10^3 + 7.2 \times 10^3 = 10 \times 10^3 \ Pa = 10^4 \ Pa$.
The mole fraction of $A$ in the vapour phase $(y_A)$ is given by $y_A = \frac{P_A}{P_T} = \frac{\chi_A P_A^o}{P_T}$.
$y_A = \frac{0.4 \times 7 \times 10^3}{10^4} = \frac{2800}{10000} = 0.28$.
The mole fraction of $B$ in the vapour phase is $y_B = 1 - y_A = 1 - 0.28 = 0.72$.

(A) Given: $P_M^o = 450 \ mmHg$ and $P_N^o = 700 \ mmHg$.
According to Raoult's law,the partial pressures are $P_M = x_M P_M^o = 450 x_M$ and $P_N = x_N P_N^o = 700 x_N$.
In the vapour phase,the mole fractions are given by $y_M = \frac{P_M}{P_T}$ and $y_N = \frac{P_N}{P_T}$,where $P_T$ is the total pressure.
Therefore,the ratio $\frac{y_M}{y_N} = \frac{P_M}{P_N} = \frac{450 x_M}{700 x_N} = \frac{450}{700} \times \frac{x_M}{x_N}$.
Since $\frac{450}{700} < 1$,it follows that $\frac{y_M}{y_N} < \frac{x_M}{x_N}$,which implies $\frac{x_M}{x_N} > \frac{y_M}{y_N}$.

(A) For an ideal mixture,the partial pressure of a component $i$ in the vapour phase is given by $P_i = P_i^o X_i$.
Given the molar ratio is $1:1$,the mole fractions are $X_{benzene} = 0.5$ and $X_{toluene} = 0.5$.
The partial pressures are $P_{benzene} = 75 \times 0.5 = 37.5 \, mm \, Hg$ and $P_{toluene} = 22 \times 0.5 = 11 \, mm \, Hg$.
The mole fraction in the vapour phase is $Y_i = P_i / P_{total}$.
Since $P_{benzene} > P_{toluene}$,it follows that $Y_{benzene} > Y_{toluene}$.
Therefore,the vapour will contain a higher percentage of benzene.

(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. $C_2H_5OH$ exhibits strong intermolecular hydrogen bonding,whereas $C_2H_5I$ does not. This difference in intermolecular forces leads to significant deviations from Raoult's law.

(B) For a solution to show positive deviation from Raoult's law,the intermolecular forces between the solute and solvent molecules ($A-B$ interaction) must be weaker than the forces between the pure components ($A-A$ and $B-B$ interactions).
This results in an increase in the total volume of the mixture $(\Delta V_{mix} > 0)$ and an absorption of heat $(\Delta H_{mix} > 0)$.

(A) Given the ratio of vapour pressures $P_A^o : P_B^o = 1 : 3$ and the ratio of mole fractions in the vapour phase $y_A : y_B = 4 : 3$.
According to Raoult's law and Dalton's law,the mole fraction in the vapour phase is given by $y_A = \frac{P_A}{P_{total}} = \frac{x_A P_A^o}{P_{total}}$ and $y_B = \frac{P_B}{P_{total}} = \frac{x_B P_B^o}{P_{total}}$.
Therefore,$\frac{y_A}{y_B} = \frac{x_A P_A^o}{x_B P_B^o}$.
Substituting the given values: $\frac{4}{3} = \frac{x_A}{x_B} \times \frac{1}{3}$.
This simplifies to $\frac{x_A}{x_B} = 4$,which means $x_A = 4x_B$.
Since the sum of mole fractions in the solution is $x_A + x_B = 1$,we substitute $x_A$: $4x_B + x_B = 1$.
$5x_B = 1$,so $x_B = \frac{1}{5}$.

(D) The mole fraction of $A$ $(X_A)$ and $B$ $(X_B)$ in the liquid phase are:
$X_A = \frac{2}{2+2} = 0.5$
$X_B = \frac{2}{2+2} = 0.5$
The total vapour pressure of the solution $(P_S)$ is calculated as:
$P_S = P_A^o X_A + P_B^o X_B$
$P_S = 120 \times 0.5 + 80 \times 0.5 = 60 + 40 = 100 \ torr$
According to Dalton's Law,the partial pressure of $B$ in the vapour phase is $P_B = P_B^o X_B = P_S Y_B$,where $Y_B$ is the mole fraction of $B$ in the vapour phase:
$80 \times 0.5 = 100 \times Y_B$
$40 = 100 \times Y_B$
$Y_B = \frac{40}{100} = \frac{2}{5}$

(B) 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 $B-B$ interactions).
This leads to:
$1$. $\Delta H_{mix} < 0$ (Exothermic process).
$2$. $\Delta V_{mix} < 0$ (Volume contraction).
Therefore,the $A-B$ interaction is stronger than $A-A$ and $B-B$ interactions.

(B) $1$. Calculate the moles of benzene $(A)$ and chlorobenzene $(B)$:
$n_A = \frac{1560}{78} = 20 \ mol$
$n_B = \frac{1125}{112.5} = 10 \ mol$
$2$. Calculate the mole fractions:
$X_A = \frac{20}{20+10} = \frac{2}{3}$
$X_B = \frac{10}{20+10} = \frac{1}{3}$
$3$. The boiling point is the temperature where the total vapor pressure $P_s = P_A^0 X_A + P_B^0 X_B = 1000 \ torr$.
$4$. Test $100 \ ^{\circ}C$: From the graph,$P_A^0 = 1350 \ torr$ and $P_B^0 = 300 \ torr$.
$P_s = (1350 \times \frac{2}{3}) + (300 \times \frac{1}{3}) = 900 + 100 = 1000 \ torr$.
Since $P_s = 1000 \ torr$ at $100 \ ^{\circ}C$,the boiling point is $100 \ ^{\circ}C$.

(A) When ethanol is mixed with acetone,the hydrogen bonding between ethanol molecules is disrupted by the addition of acetone. This leads to a weaker interaction between the components compared to the pure substances. Consequently,the solution shows a positive deviation from Raoult's law,which is characterized by an endothermic process where $\Delta H_{mix} > 0$. Therefore,heat is absorbed.

(B) $1$. Calculate the moles of each component:
$n_{\text{benzene}} = \frac{1560}{78} = 20 \ mol$
$n_{\text{chlorobenzene}} = \frac{1125}{112.5} = 10 \ mol$
$2$. Calculate the mole fractions:
$\chi_{\text{benzene}} = \frac{20}{20+10} = \frac{2}{3}$
$\chi_{\text{chlorobenzene}} = \frac{10}{20+10} = \frac{1}{3}$
$3$. The boiling point is the temperature where the total vapor pressure $P_{\text{total}} = P_{\text{benzene}}^0 \chi_{\text{benzene}} + P_{\text{chlorobenzene}}^0 \chi_{\text{chlorobenzene}} = 1000 \ torr$.
$4$. Testing $100^{\circ}C$ from the graph:
$P_{\text{benzene}}^0 = 1350 \ torr$
$P_{\text{chlorobenzene}}^0 = 300 \ torr$
$P_{\text{total}} = (1350 \times \frac{2}{3}) + (300 \times \frac{1}{3}) = 900 + 100 = 1000 \ torr$.
$5$. Since the total pressure equals the external pressure at $100^{\circ}C$,the boiling point is $100^{\circ}C$.

(B) For an ideal solution,the partial pressure of component $A$ is given by $P_A = X_A P_A^{\circ}$ and the total pressure is $P_{total} = X_A P_A^{\circ} + X_B P_B^{\circ}$.
According to Dalton's law of partial pressures,$P_A = Y_A P_{total}$.
Equating the two expressions for $P_A$: $Y_A P_{total} = X_A P_A^{\circ} \Rightarrow \frac{Y_A}{X_A} = \frac{P_A^{\circ}}{P_{total}}$.
Since $P_A^{\circ} > P_B^{\circ}$,the component with the higher vapour pressure will be more concentrated in the vapour phase than in the liquid phase.
Therefore,$Y_A > X_A$,which implies $\frac{Y_A}{X_A} > 1$.

(B) For an ideal solution,the enthalpy of mixing is zero $(\Delta H_{mix} = 0)$ and the volume of mixing is zero $(\Delta V_{mix} = 0)$.
However,for the formation of a solution,the entropy of mixing is always positive $(\Delta S_{mix} > 0)$ and the Gibbs free energy of mixing is always negative $(\Delta G_{mix} < 0)$.

(B) Given: $P_A^{\circ} = 300 \, torr$,$P_B^{\circ} = 800 \, torr$.
Mole fraction of $B$ $(x_B)$ = $0.92$,so $x_A = 1 - 0.92 = 0.08$.
According to Raoult's Law,the expected total pressure $(P_{ideal})$ is:
$P_{ideal} = P_A^{\circ} x_A + P_B^{\circ} x_B = (300 \times 0.08) + (800 \times 0.92) = 24 + 736 = 760 \, torr$.
Since $1 \, atm = 760 \, torr$,the observed pressure is $0.95 \, atm = 0.95 \times 760 = 722 \, torr$.
Since the observed pressure $(722 \, torr)$ is less than the ideal pressure $(760 \, torr)$,the solution shows a negative deviation from Raoult's Law.
For solutions showing negative deviation,$\Delta H_{mix} < 0$ and $\Delta V_{mix} < 0$.

(C) Raoult's law is obeyed by an ideal solution.
In an ideal solution,the intermolecular forces of attraction between the components ($A-A$ and $B-B$) are identical to the intermolecular forces of attraction between the unlike components $(A-B)$.
Therefore,the correct condition is that the forces of attraction between like molecules are identical to those between unlike molecules.

(D) An ideal solution is formed when the solute-solute and solvent-solvent interactions are similar to the solute-solvent interactions.
$C_6H_6$ (benzene) and $C_6H_5CH_3$ (toluene) have similar molecular structures and polarities,resulting in nearly identical intermolecular forces.
Therefore,they form an ideal solution over the entire range of composition.
Other options represent non-ideal solutions showing either positive or negative deviations from Raoult's law.

(C) 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).
$4$. $\Delta G_{mix} < 0$ (Free energy of mixing is negative,as the process is spontaneous).
Therefore,option $C$ is the correct statement.

(B) According to Raoult's Law,the total pressure of a binary ideal solution is given by $P_{total} = X_A P_A^o + X_B P_B^o$.
Since $X_A + X_B = 1$,we can substitute $X_A = 1 - X_B$.
$P_{total} = (1 - X_B) P_A^o + X_B P_B^o$
$P_{total} = P_A^o - X_B P_A^o + X_B P_B^o$
$P_{total} = P_A^o + X_B(P_B^o - P_A^o)$.
Alternatively,substituting $X_B = 1 - X_A$:
$P_{total} = X_A P_A^o + (1 - X_A) P_B^o$
$P_{total} = X_A P_A^o + P_B^o - X_A P_B^o$
$P_{total} = P_B^o + X_A(P_A^o - P_B^o)$.
Comparing this with the given options,option $B$ is correct.

(A) According to Raoult's Law for an ideal solution,the total vapor pressure $P_{total} = P_A^o x_A + P_B^o x_B$.
Given the mole ratio $n_A : n_B = 3 : 1$,the mole fractions are $x_A = \frac{3}{3+1} = 0.75$ and $x_B = \frac{1}{3+1} = 0.25$.
Substituting the given values: $P_{total} = (24 \ torr \times 0.75) + (40 \ torr \times 0.25)$.
$P_{total} = 18 \ torr + 10 \ torr = 28 \ torr$.

(D) According to Raoult's Law,the partial vapor pressure of a component in an ideal solution is given by $p_i = x_i p_i^0$,where $x_i$ is the mole fraction and $p_i^0$ is the vapor pressure of the pure component.
However,the question provides the partial vapor pressures ($p_B = 1.55 \ kPa$ and $p_T = 1.85 \ kPa$) directly.
In an ideal solution,the ratio of partial pressures is equal to the ratio of the products of mole fractions and pure vapor pressures.
Since $p_B = x_B p_B^0$ and $p_T = x_T p_T^0$,the ratio of partial pressures is $\frac{p_B}{p_T} = \frac{x_B p_B^0}{x_T p_T^0}$.
Without the pure vapor pressures ($p_B^0$ and $p_T^0$),the mole ratio $\frac{x_B}{x_T}$ cannot be calculated from the partial pressures alone.
Therefore,the correct answer is that it cannot be determined.

(B) An ideal solution is formed by components that have similar molecular structures and polarities,resulting in similar intermolecular forces of attraction between $A-A$,$B-B$,and $A-B$ molecules.
Benzene $(C_6H_6)$ and Toluene $(C_6H_5CH_3)$ have very similar structures and chemical properties.
When mixed,they form an ideal solution because the intermolecular forces in the mixture are nearly identical to those in the pure components.
$CHCl_3$ and $CH_3COCH_3$ show negative deviation due to hydrogen bonding.
$CH_3CH_2OH$ and $H_2O$ show positive deviation.
$H_2SO_4$ and $H_2O$ show large negative deviation due to strong exothermic interaction.

(B) Given: $n_A = 1 \, \text{mole}$,$n_B = 2 \, \text{mole}$.
Total moles $n_T = 1 + 2 = 3 \, \text{mole}$.
Mole fractions: $x_A = 1/3$,$x_B = 2/3$.
Pure vapor pressures: $P_A^0 = 45 \, \text{torr}$,$P_B^0 = 36 \, \text{torr}$.
According to Raoult's Law,the expected vapor pressure $P_{\text{ideal}} = P_A^0 x_A + P_B^0 x_B$.
$P_{\text{ideal}} = (45 \times 1/3) + (36 \times 2/3) = 15 + 24 = 39 \, \text{torr}$.
The observed vapor pressure is $38 \, \text{torr}$.
Since $P_{\text{observed}} < P_{\text{ideal}}$ $(38 < 39)$,the solution shows negative deviation from Raoult's Law.

(B) positive deviation from Raoult's law occurs when the solute-solvent intermolecular forces are weaker than the solute-solute and solvent-solvent forces.
In the mixture of $Benzene$ and $Methanol$,the hydrogen bonding between $Methanol$ molecules is disrupted by the addition of $Benzene$,leading to weaker interactions and a positive deviation.
$Water-HCl$ and $Water-Nitric acid$ show negative deviations due to strong hydrogen bonding formation.
$Acetone-Chloroform$ shows a negative deviation due to the formation of intermolecular hydrogen bonds between the two components.

(B) negative deviation from Raoult's law occurs when the intermolecular forces of attraction between the solute and solvent molecules are stronger than those between the pure components ($A-A$ or $B-B$ interactions).
In the mixture of $Acetone$ $(CH_3COCH_3)$ and $Chloroform$ $(CHCl_3)$,a strong hydrogen bond is formed between the oxygen atom of acetone and the hydrogen atom of chloroform.
This increased attraction leads to a decrease in the total vapor pressure of the solution compared to the expected value,which is characteristic of a negative deviation.

(A) Acetone and ethanol form a non-ideal solution showing positive deviation from Raoult's law.
In such solutions,the intermolecular forces of attraction between $A-B$ (acetone-ethanol) are weaker than those between $A-A$ (acetone-acetone) and $B-B$ (ethanol-ethanol).
Due to weaker interactions,the molecules move further apart,leading to an increase in the total volume of the mixture.
Therefore,the final volume will be greater than the sum of the individual volumes,i.e.,$> 50 \ mL$.

(D) An ideal solution follows Raoult's Law,which states that the partial pressure of each component is directly proportional to its mole fraction: $P_A = P_A^0 X_A$ and $P_B = P_B^0 X_B$.
Therefore,the plots of $P_A$ vs $X_A$ and $P_B$ vs $X_B$ are straight lines passing through the origin.
The total pressure is given by $P_{total} = P_A + P_B = P_A^0 X_A + P_B^0 (1 - X_A) = (P_A^0 - P_B^0) X_A + P_B^0$.
This equation represents a straight line with respect to $X_A$ (or $X_B$).
Thus,the statement that the plot of $P_{total}$ vs $X_A$ is not a straight line is incorrect for an ideal solution.

(D) Given: $x_A = 0.7$,$x_B = 1 - 0.7 = 0.3$,$y_A = 0.4$,$y_B = 1 - 0.4 = 0.6$.
According to Raoult's law,$P_A = x_A P_A^o$ and $P_B = x_B P_B^o$.
The total pressure $P_T = P_A + P_B = x_A P_A^o + x_B P_B^o = 0.7 P_A^o + 0.3 P_B^o$.
Also,for vapor phase,$y_A = \frac{P_A}{P_T} = \frac{x_A P_A^o}{P_T}$.
So,$P_T = \frac{x_A P_A^o}{y_A} = \frac{0.7 P_A^o}{0.4} = 1.75 P_A^o$.
Equating the two expressions for $P_T$: $1.75 P_A^o = 0.7 P_A^o + 0.3 P_B^o$.
$1.05 P_A^o = 0.3 P_B^o \implies P_B^o = \frac{1.05}{0.3} P_A^o = 3.5 P_A^o$.
Given $P_A^o + P_B^o = 90 \, mm$,substitute $P_B^o$: $P_A^o + 3.5 P_A^o = 90 \implies 4.5 P_A^o = 90$.
$P_A^o = 20 \, mm$ and $P_B^o = 3.5 \times 20 = 70 \, mm$.

(A) Given: Mole fraction of $B$ in liquid phase $(x_B)$ = $0.4$,Mole fraction of $B$ in vapor phase $(y_B)$ = $0.25$,$P_B^o = 40 \ mm \ Hg$.
Since $x_A + x_B = 1$,$x_A = 1 - 0.4 = 0.6$.
According to Raoult's Law,the total pressure $P_T = P_A^o x_A + P_B^o x_B = P_A^o(0.6) + 40(0.4) = 0.6 P_A^o + 16$.
Also,the partial pressure of $B$ is $P_B = P_T y_B$.
$P_B = P_B^o x_B = 40 \times 0.4 = 16 \ mm \ Hg$.
Substituting $P_B$ in the vapor phase equation: $16 = P_T \times 0.25$,so $P_T = 16 / 0.25 = 64 \ mm \ Hg$.
Now,$64 = 0.6 P_A^o + 16$.
$0.6 P_A^o = 64 - 16 = 48$.
$P_A^o = 48 / 0.6 = 80 \ mm \ Hg$.

(B) An aqueous solution of $HCl$ shows a negative deviation from Raoult's law.
In this solution,the intermolecular forces of attraction between the solute $(HCl)$ and solvent $(H_2O)$ molecules are stronger than the forces between the pure components.
This stronger attraction reduces the escaping tendency of the molecules into the vapour phase.
Consequently,the observed partial vapour pressure is lower than the value predicted by Raoult's law,which is the characteristic definition of a negative deviation.

(A) According to Raoult's law for an ideal solution,the total vapour pressure $P_{total}$ is given by:
$P_{total} = P_A^o x_A + P_B^o x_B$
Given: $P_A^o = 120 \, mm \, Hg$,$P_B^o = 180 \, mm \, Hg$,$n_A = 2 \, moles$,$n_B = 3 \, moles$.
Total moles $= 2 + 3 = 5 \, moles$.
Mole fraction of $A$ $(x_A)$ $= \frac{2}{5} = 0.4$.
Mole fraction of $B$ $(x_B)$ $= \frac{3}{5} = 0.6$.
$P_{total} = (120 \times 0.4) + (180 \times 0.6) = 48 + 108 = 156 \, mm \, Hg$.