Ideal and Non-ideal solution Questions in English

(D) An ideal solution is formed by components with similar molecular structures and polarities,such as $n-hexane$ and $n-heptane$.
$C_2H_5OH + H_2O$ shows positive deviation due to the disruption of hydrogen bonding.
$Acetone + Chloroform$ shows negative deviation from Raoult's Law because of the formation of strong intermolecular hydrogen bonding between them.
$Chloroform + Benzene$ is not an ideal solution; it shows negative deviation from Raoult's Law. Thus,option $D$ is incorrect.

(B) Ideal solutions are those that obey Raoult's law at all temperatures and concentrations.
In these solutions,the solute-solute,solvent-solvent,and solute-solvent interactions are nearly identical.
$I$. Chloroethane and bromoethane have similar structures and polarities,forming an ideal solution.
$II$. Benzene and toluene have similar structures and intermolecular forces,forming an ideal solution.
$III$. $n$-hexane and $n$-heptane are homologous alkanes with similar intermolecular forces,forming an ideal solution.
$IV$. Phenol and aniline form a non-ideal solution showing negative deviation from Raoult's law due to strong hydrogen bonding between phenol and aniline molecules,resulting in $\Delta H_{mix} < 0$ and $\Delta V_{mix} < 0$.
Therefore,$I, II$ and $III$ form an ideal solution.

(A) Ideal solutions are formed when the solute and solvent molecules have similar sizes,identical polarity,and comparable intermolecular forces.
Ethyl chloride and ethyl bromide form an ideal solution because their molecular structures and intermolecular interactions are very similar.
For an ideal solution,the enthalpy change of mixing is $\Delta_{\text{mix}}H = 0$ and the volume change of mixing is $\Delta_{\text{mix}}V = 0$.

(D) An ideal solution is one that obeys Raoult's law over the entire range of concentration and temperature.
For an ideal solution,the enthalpy of mixing $(\Delta H_{mix})$ is $0$ and the volume of mixing $(\Delta V_{mix})$ is $0$.
Benzene $(C_6H_6)$ and toluene $(C_6H_5CH_3)$ have similar structures and intermolecular forces,which leads to the formation of an ideal solution.
Other options like $C_2H_5OH \& H_2O$,$HNO_3 \& H_2O$,and $CHCl_3 \& CH_3COCH_3$ show deviations from Raoult's law due to differences in intermolecular interactions.
Therefore,$C_6H_6 \& C_6H_5CH_3$ form an ideal solution.

(A) 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 change of mixing is zero $(\Delta_{mix} H = 0)$ and the volume change of mixing is zero $(\Delta_{mix} V = 0)$.
However,when two components are mixed to form a solution,the entropy of the system increases due to the increase in randomness,so $\Delta_{mix} S > 0$.
Therefore,the condition $\Delta_{mix} S = 0$ is not satisfied by an ideal solution.

(C) An ideal solution obeys Raoult’s law at all temperatures and pressures.
In an ideal solution,the solute-solute and solvent-solvent interactions are similar to the solute-solvent interactions.
$1$. Benzene + Toluene: Ideal solution.
$2$. Chlorobenzene + $1, 2-$dichlorobenzene: Ideal solution.
$3$. Methyl iodide + Isopropanol: Non-ideal solution (due to different intermolecular forces).
$4$. Ethyl bromide + Methyl bromide: Ideal solution.
Therefore,the pair of methyl iodide and isopropanol does not form an ideal solution.

(B) $1$. Calculate moles of each component:
$n_{CHCl_3} = \frac{59.75 \ g}{119.5 \ g \cdot mol^{-1}} = 0.5 \ mol$
$n_{CH_2Cl_2} = \frac{21.25 \ g}{85 \ g \cdot mol^{-1}} = 0.25 \ mol$
$2$. Calculate mole fractions in liquid phase:
$x_{CHCl_3} = \frac{0.5}{0.5 + 0.25} = \frac{0.5}{0.75} = \frac{2}{3}$
$x_{CH_2Cl_2} = \frac{0.25}{0.5 + 0.25} = \frac{0.25}{0.75} = \frac{1}{3}$
$3$. Calculate partial pressures:
$P_{CHCl_3} = x_{CHCl_3} \times P^0_{CHCl_3} = \frac{2}{3} \times 200 \ mmHg = 133.33 \ mmHg$
$P_{CH_2Cl_2} = x_{CH_2Cl_2} \times P^0_{CH_2Cl_2} = \frac{1}{3} \times 415 \ mmHg = 138.33 \ mmHg$
$4$. Calculate total pressure:
$P_{total} = 133.33 + 138.33 = 271.66 \ mmHg$
$5$. Calculate mole fractions in vapour phase $(y_i = \frac{P_i}{P_{total}})$:
$y_{CHCl_3} = \frac{133.33}{271.66} \approx 0.491$
$y_{CH_2Cl_2} = \frac{138.33}{271.66} \approx 0.509$

(B) Given:
$P_{B}^{\circ} = 50 \ mmHg$,$P_{T}^{\circ} = 40 \ mmHg$
$M_{B} = 78 \ g \ mol^{-1}$,$M_{T} = 92 \ g \ mol^{-1}$
Moles of benzene $(n_{B})$ $= \frac{117 \ g}{78 \ g \ mol^{-1}} = 1.5 \ mol$
Moles of toluene $(n_{T})$ $= \frac{46 \ g}{92 \ g \ mol^{-1}} = 0.5 \ mol$
Mole fraction of benzene $(X_{B})$ $= \frac{1.5}{1.5 + 0.5} = 0.75$
Mole fraction of toluene $(X_{T})$ $= \frac{0.5}{1.5 + 0.5} = 0.25$
Partial pressure of benzene $(P_{B})$ $= P_{B}^{\circ} \times X_{B} = 50 \times 0.75 = 37.5 \ mmHg$
Partial pressure of toluene $(P_{T})$ $= P_{T}^{\circ} \times X_{T} = 40 \times 0.25 = 10 \ mmHg$
Total vapour pressure $(P_{total})$ $= P_{B} + P_{T} = 37.5 + 10 = 47.5 \ mmHg$
Mole fraction of toluene in vapour phase $(Y_{T})$ $= \frac{P_{T}}{P_{total}} = \frac{10}{47.5} \approx 0.21$

(C) For an ideal solution,the total vapour pressure $P_T$ is given by Raoult's law: $P_T = P_A^0 x_A + P_B^0 x_B$,where $x_A + x_B = 1$.
Case $1$: Molar ratio $1:1$,so $x_A = 0.5$ and $x_B = 0.5$. $P_T = 400 \ mm$. Thus,$0.5 P_A^0 + 0.5 P_B^0 = 400$,which simplifies to $P_A^0 + P_B^0 = 800$ (Equation $1$).
Case $2$: Molar ratio $1:2$,so $x_A = 1/3$ and $x_B = 2/3$. $P_T = 350 \ mm$. Thus,$(1/3) P_A^0 + (2/3) P_B^0 = 350$,which simplifies to $P_A^0 + 2 P_B^0 = 1050$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(P_A^0 + 2 P_B^0) - (P_A^0 + P_B^0) = 1050 - 800$,so $P_B^0 = 250 \ mm$.
Substituting $P_B^0 = 250$ into Equation $1$: $P_A^0 + 250 = 800$,so $P_A^0 = 550 \ mm$.
Therefore,the vapour pressures of pure liquids $A$ and $B$ are $550 \ mm$ and $250 \ mm$ respectively.

(A) According to Raoult's law for an ideal solution,the total vapour pressure $P_{total}$ is given by the sum of the partial pressures of the components: $P_{total} = P_A + P_B$.
Given:
Mole fraction of $A$ $(x_A)$ = $0.67$
Mole fraction of $B$ $(x_B)$ = $0.33$
Vapour pressure of pure $A$ $(P^{\circ}_A)$ = $300 \ mm$
Vapour pressure of pure $B$ $(P^{\circ}_B)$ = $450 \ mm$
Using the formula $P_{total} = x_A P^{\circ}_A + x_B P^{\circ}_B$:
$P_{total} = (0.67 \times 300) + (0.33 \times 450)$
$P_{total} = 201 + 148.5 = 349.5 \ mm$.

(A) $1$. Calculate the molar masses: Heptane $(C_7H_{16})$ = $7 \times 12 + 16 \times 1 = 100 \ g/mol$. Octane $(C_8H_{18})$ = $8 \times 12 + 18 \times 1 = 114 \ g/mol$.
$2$. Calculate the number of moles: $n_{\text{heptane}} = \frac{25 \ g}{100 \ g/mol} = 0.25 \ mol$. $n_{\text{octane}} = \frac{57 \ g}{114 \ g/mol} = 0.50 \ mol$.
$3$. Calculate the mole fractions: $x_{\text{heptane}} = \frac{0.25}{0.25 + 0.50} = \frac{0.25}{0.75} = \frac{1}{3}$. $x_{\text{octane}} = \frac{0.50}{0.75} = \frac{2}{3}$.
$4$. Apply Raoult's Law: $P_{\text{total}} = P^{\circ}_{\text{heptane}} \times x_{\text{heptane}} + P^{\circ}_{\text{octane}} \times x_{\text{octane}}$.
$5$. $P_{\text{total}} = 106 \times \frac{1}{3} + 47 \times \frac{2}{3} = \frac{106 + 94}{3} = \frac{200}{3} = 66.66 \ kPa$.

(D) Given: $P_A^\circ = 500 \ mm \ Hg$,$P_B^\circ = 400 \ mm \ Hg$.
Since equal moles are mixed,$x_A = x_B = 0.5$.
Total pressure $P_{total} = P_A^\circ x_A + P_B^\circ x_B = (500 \times 0.5) + (400 \times 0.5) = 250 + 200 = 450 \ mm \ Hg$.
Mole fraction in vapor phase: $y_A = \frac{P_A}{P_{total}} = \frac{250}{450} = 0.555$ and $y_B = \frac{P_B}{P_{total}} = \frac{200}{450} = 0.444$.

(C) According to Raoult's law,the total vapour pressure of an ideal solution is given by $P_{total} = X_A P_A^o + X_B P_B^o$.
For the first case: $n_A = 1, n_B = 3$. Total moles = $4$. $X_A = 1/4, X_B = 3/4$.
$550 = (1/4)P_A^o + (3/4)P_B^o \implies P_A^o + 3P_B^o = 2200$ (Equation $1$).
For the second case: $n_A = 1, n_B = 4$. Total moles = $5$. $X_A = 1/5, X_B = 4/5$.
$560 = (1/5)P_A^o + (4/5)P_B^o \implies P_A^o + 4P_B^o = 2800$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(P_A^o + 4P_B^o) - (P_A^o + 3P_B^o) = 2800 - 2200 \implies P_B^o = 600 \ mm \ Hg$.
Substituting $P_B^o$ in Equation $1$: $P_A^o + 3(600) = 2200 \implies P_A^o + 1800 = 2200 \implies P_A^o = 400 \ mm \ Hg$.
The ratio $P_A^o : P_B^o = 400 : 600 = 2 : 3$.

(D) Given: $P_A^0 = 50 \ mm \ Hg$,$P_B^0 = 32 \ mm \ Hg$.
Mole fraction in liquid phase: $X_A = \frac{1}{1+1} = 0.5$ and $X_B = \frac{1}{1+1} = 0.5$.
Partial pressures: $P_A = P_A^0 \times X_A = 50 \times 0.5 = 25 \ mm \ Hg$ and $P_B = P_B^0 \times X_B = 32 \times 0.5 = 16 \ mm \ Hg$.
Total pressure: $P_{total} = P_A + P_B = 25 + 16 = 41 \ mm \ Hg$.
Mole fraction of $A$ in vapour phase $(Y_A)$: $Y_A = \frac{P_A}{P_{total}} = \frac{25}{41} \approx 0.6097 \approx 0.61$.

(C) An ideal solution is a solution that obeys Raoult's law over the entire range of concentration.
Such solutions are formed by mixing two components that are similar in molecular size,structure,and have almost identical intermolecular forces.
The characteristics of ideal solutions are:
$(1)$ It must obey Raoult's law.
$(2)$ The enthalpy of mixing should be zero,i.e.,$\Delta H_{\text{mix}} = 0$.
$(3)$ The volume of mixing should be zero,i.e.,$\Delta V_{\text{mix}} = 0$,which implies $V_{\text{solvent}} + V_{\text{solute}} = V_{\text{solution}}$.
Examples of ideal solutions include $n$-hexane + $n$-heptane and benzene + toluene.
The mixture of carbon dioxide and water is not an ideal solution because they react with each other to form $H_2CO_3$,which involves chemical changes and significant enthalpy changes.
Therefore,statements $(a)$,$(b)$,and $(c)$ are correct.

(D) The mixture of $CH_3OH$ (methanol) and $CH_3COOH$ (acetic acid) forms a non-ideal solution showing negative deviation from Raoult's law due to strong hydrogen bonding between the components.
For solutions showing negative deviation:
$1$. $\Delta H_{\text{mix}} < 0$ (Exothermic process).
$2$. $\Delta V_{\text{mix}} < 0$.
$3$. They do not obey Raoult's law.
Evaluating the given statements:
$(a)$ $\Delta H_{\text{mix}} < 0$ is correct.
$(b)$ Does not obey Raoult's law is correct.
$(c)$ $\Delta H_{\text{mix}} > 0$ is incorrect.
$(d)$ It is not an ideal solution,so it is incorrect.
Therefore,the statements that are "not correct" are $(c)$ and $(d)$.

(C) An ideal solution is defined as a solution that obeys Raoult's law over the entire range of concentrations.
For an ideal solution,the enthalpy of mixing $(\Delta H_{\text{mix}})$ is $0$ and the volume of mixing $(\Delta V_{\text{mix}})$ is $0$.
This occurs because the intermolecular forces of attraction between solute-solute,solvent-solvent,and solute-solvent particles are nearly identical.
Therefore,statements $(II)$ and $(III)$ are correct.

(B) Solution $A$ (Phenol and aniline) shows negative deviation from Raoult's law because the intermolecular hydrogen bonding between phenol and aniline molecules is stronger than the individual $A-A$ and $B-B$ interactions.
Solution $B$ (Chloroform and acetone) also shows negative deviation because the formation of hydrogen bonding between the chloroform $(CHCl_3)$ and acetone $(CH_3COCH_3)$ molecules makes the $A-B$ interaction stronger than the $A-A$ and $B-B$ interactions.
Therefore,both solutions $A$ and $B$ show $-ve$ deviation.

(D) An ideal solution obeys Raoult's law at all temperatures and pressures.
In an ideal solution,the solute-solute and solvent-solvent interactions are similar to the solute-solvent interactions.
There is no change in enthalpy $( \Delta H_{mix} = 0 )$ and no change in volume $( \Delta V_{mix} = 0 )$ upon mixing.
$H_2O + CH_3OH$ shows positive deviation from Raoult's law.
$CHCl_3 + C_6H_6$ shows negative deviation from Raoult's law.
$CCl_4 + C_7H_8$ is also non-ideal.
$n-C_6H_{14} + n-C_7H_{16}$ (n-hexane and n-heptane) form an ideal solution because they have similar molecular structures and polarities.

(A) For an ideal solution,the following conditions must be satisfied:
$1$. $\Delta_{mix} H = 0$: There is no heat absorbed or evolved during the mixing process.
$2$. $\Delta_{mix} V = 0$: The total volume of the solution is equal to the sum of the volumes of the individual components.
$3$. $\Delta_{mix} S > 0$: The entropy of mixing is positive because the mixing process increases the randomness of the system.
$4$. $\Delta_{mix} G < 0$: The Gibbs free energy of mixing is negative,which makes the formation of an ideal solution a spontaneous process.
Since all four conditions $(A, B, C, D)$ are correct for an ideal solution,the correct option is $(A)$.

(D) For real solutions showing negative deviation from Raoult's law,the solute-solvent interactions are stronger than the solute-solute or solvent-solvent interactions.
This leads to a decrease in the total volume of the solution,meaning $\Delta V_{\text{mix}} < 0$.
Additionally,the formation of stronger bonds releases energy,making the process exothermic,which means $\Delta H_{\text{mix}} < 0$.

(C) For the formation of an ideal solution,the process is spontaneous,which implies $\Delta G < 0$.
According to the definition of an ideal solution:
$\Delta H_{\text{mix}} = 0$
$\Delta V_{\text{mix}} = 0$
Since $\Delta H_{\text{mix}} = 0$,there is no heat exchange between the system and the surroundings,so $\Delta S_{\text{surroundings}} = 0$.
Mixing increases the randomness of the components,so $\Delta S_{\text{system}} > 0$.
Using the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$.
Since $\Delta H = 0$ and $\Delta S > 0$,we get $\Delta G = -T\Delta S$,which is less than $0$ (negative).

(D) When $Acetone$ $(CH_3COCH_3)$ and $Chloroform$ $(CHCl_3)$ are mixed,they form intermolecular hydrogen bonds between the oxygen atom of acetone and the hydrogen atom of chloroform.
This interaction is stronger than the original $Acetone-Acetone$ and $Chloroform-Chloroform$ interactions.
As a result,the escaping tendency of the molecules decreases,leading to a negative deviation from Raoult's law.

(A) According to Raoult's law and Dalton's law,the relationship between mole fraction in vapour phase $(Y_A)$ and liquid phase $(X_A)$ is given by:
$Y_A = \frac{P_A^0 X_A}{P_{total}}$
Since $P_{total} = P_A^0 X_A + P_B^0 X_B$,we have:
$Y_A = \frac{P_A^0 X_A}{P_A^0 X_A + P_B^0 (1 - X_A)}$
Given $Y_A = 0.8$,$P_A^0 = 55$,and $P_B^0 = 15$:
$0.8 = \frac{55 X_A}{55 X_A + 15(1 - X_A)}$
$0.8 = \frac{55 X_A}{40 X_A + 15}$
$32 X_A + 12 = 55 X_A$
$23 X_A = 12$
$X_A = \frac{12}{23} \approx 0.5217$

(A) mixture of $CS_{2}$ and acetone exhibits a positive deviation from Raoult's law.
In a positive deviation,the partial pressure of each component is greater than that predicted by Raoult's law.
Mathematically,this is represented as $P_{CS_{2}} > P_{CS_{2}}^{o} \cdot X_{CS_{2}}$.
Among the given options,the graph that shows the partial pressure curve for $CS_{2}$ lying above the ideal straight line (Raoult's law) is the correct representation.

(C) Mixtures that exhibit $\Delta_{mix} H > 0$ show a positive deviation from Raoult's law.
In such mixtures,the solute-solvent interactions are weaker than the solute-solute and solvent-solvent interactions present in the pure components.
This results in the absorption of heat upon mixing,making the process endothermic $(\Delta_{mix} H > 0)$.
Option $(A)$ $(CHCl_3 + CH_3COCH_3)$ shows negative deviation due to strong hydrogen bonding.
Option $(B)$ $(C_6H_5NH_2 + C_6H_5OH)$ shows negative deviation due to acid-base interaction.
Option $(C)$ $(C_2H_5OH + CH_3COCH_3)$ shows positive deviation because the strong hydrogen bonds in pure ethanol are disrupted when acetone is added,and the new interactions are weaker.
Option $(D)$ $(C_6H_6 + C_6H_5CH_3)$ forms an ideal solution with $\Delta_{mix} H \approx 0$.
Therefore,the correct option is $(C)$.

(A) Negative deviation from Raoult's law occurs when the intermolecular forces between the solute and solvent molecules are stronger than the forces between the pure components (solute-solute and solvent-solvent interactions).
When Chloroform $(CHCl_3)$ is mixed with Acetone $(CH_3COCH_3)$,a strong hydrogen bond is formed between the oxygen atom of the acetone and the hydrogen atom of the chloroform.
This increased intermolecular attraction results in a decrease in total volume and vapor pressure,which is characteristic of negative deviation.

(B) 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 is zero $(\Delta_{\text{mix}}H = 0)$,meaning no heat is absorbed or evolved during the mixing process.
Additionally,the volume of mixing is zero $(\Delta_{\text{mix}}V = 0)$,meaning the total volume of the solution is equal to the sum of the volumes of the individual components.
Therefore,the correct condition is $\Delta_{\text{mix}}H = 0$ and $\Delta_{\text{mix}}V = 0$.

(B) Chloroform $(CHCl_3)$ and acetone $(CH_3COCH_3)$ form a solution that shows a negative deviation from Raoult's law.
This occurs because the intermolecular forces,specifically the formation of hydrogen bonding between chloroform and acetone molecules,are stronger than the individual intermolecular forces present in the pure components.
This increased attraction reduces the escaping tendency of the molecules,thereby decreasing the total vapour pressure of the solution compared to the ideal behavior.