Answer

(a) According to Henry’s law partial pressure of gas in the solution is proportional to the mole fraction of gas in the solution.

where,

$$ \begin{aligned} p & =K_{H} x \\ K_{H} & =\text { Henry’s constant } \end{aligned} $$

Hence, (a) mole fraction is the correct choice.

• Parts per million (ppm): This unit is typically used to express very dilute concentrations of substances. It is not directly related to the vapor pressure of a solution, as it does not provide a direct relationship between the concentration of solute and the vapor pressure.

• Mass percentage: This unit expresses the concentration of a solution in terms of the mass of solute per mass of solution. While it is useful for various calculations, it does not directly relate to the vapor pressure of a solution, as vapor pressure is more closely related to the mole fraction of the components.

• Molality: This unit expresses the concentration of a solution in terms of the moles of solute per kilogram of solvent. Although it is useful for colligative properties like boiling point elevation and freezing point depression, it does not directly relate to the vapor pressure of a solution, which is more accurately described by the mole fraction of the solute and solvent.

Thinking Process

Use the concept of solubility and effect of temperature on solubility to answer this question.

Answer

(d) Dissolution of sugar in water will be most rapid when powdered sugar is dissolved in hot water because powder form can easily insert in the vacancies of liquid particles.

Further dissolution of sugar in water is an endothermic process. Hence, high temperature will favour the dissolution of sugar in water.

• (a) Sugar crystals in cold water: The dissolution process is slower because the sugar crystals have a smaller surface area compared to powdered sugar, and the cold water provides less kinetic energy to facilitate the dissolution process.

• (b) Sugar crystals in hot water: Although the hot water provides more kinetic energy, the larger surface area of the sugar crystals compared to powdered sugar makes the dissolution process slower.

• (c) Powdered sugar in cold water: While the powdered sugar has a larger surface area which aids in dissolution, the cold water provides less kinetic energy, making the process slower compared to using hot water.

Answer

(c) At equilibrium the rate of dissolution of solid in a volatile liquid solvent is equal to the rate of crystallisation.

• (a) The rate of dissolution being less than the rate of crystallisation would imply that the solute is precipitating out of the solution, which is not characteristic of an equilibrium state.
• (b) The rate of dissolution being greater than the rate of crystallisation would imply that the solute is continuously dissolving into the solvent, leading to a non-equilibrium state where the concentration of the solute in the solution increases over time.
• (d) A rate of zero for dissolution would mean that no solute is dissolving into the solvent, which contradicts the dynamic nature of equilibrium where both dissolution and crystallisation occur at equal rates.

Thinking Process

This problem includes concept of saturated, unsaturated, supersaturated and concentrated solution.

Answer

(b) When solute is added to the solution three cases may arise

(i) It dissolves into solution then solution is unsaturated.

(ii) It does not dissolve in the solution then solution is known as saturated.

(iii) When solute get precipitated solution is known as supersaturated solution.

• (a) Saturated: A saturated solution is one in which the maximum amount of solute has been dissolved at a given temperature. If more solute is added, it will not dissolve but will remain in the solution without precipitating. In the given scenario, the solute precipitates, indicating that the solution is beyond saturation.

• (c) Unsaturated: An unsaturated solution is one that can still dissolve more solute. If additional solute is added to an unsaturated solution, it will dissolve without any precipitation. In the given scenario, the solute precipitates, indicating that the solution cannot dissolve any more solute.

• (d) Concentrated: A concentrated solution contains a large amount of solute relative to the solvent, but it does not necessarily mean that the solution is at or beyond its saturation point. Precipitation occurs only in a supersaturated solution, not merely a concentrated one.

Answer

(c) Maximum amount of solid that can be dissolved in a specified amount of a given solvent does not depend upon pressure. This is because solid and liquid are highly incompressible and practically remain unaffected by change in pressure.

• Temperature: The solubility of a solid solute in a liquid solvent generally increases with an increase in temperature. This is because higher temperatures provide more kinetic energy to the molecules, allowing them to interact more effectively and dissolve more solute.

• Nature of solute: The solubility of a solute in a solvent depends on the chemical nature of the solute. Different solutes have different solubility properties based on their molecular structure, polarity, and other chemical characteristics.

• Nature of solvent: The solubility of a solute is also influenced by the nature of the solvent. Different solvents have different abilities to dissolve solutes based on their polarity, hydrogen bonding capacity, and other chemical properties.

Answer

(b) Low concentration of oxygen in the blood and tissues of people living at high altitude is due to low atmospheric pressure. Because at high altitude, the partial pressure of oxygen is less than at the ground level. This decreased atmospheric pressure causes release of oxygen from blood.

• (a) Low temperature: While low temperatures can affect the body’s metabolism and overall function, they do not directly cause a low concentration of oxygen in the blood and tissues. The primary factor is the reduced partial pressure of oxygen at high altitudes, not the temperature.

• (c) High atmospheric pressure: High atmospheric pressure is not a characteristic of high altitudes; in fact, it is the opposite. High atmospheric pressure is found at lower altitudes and would actually increase the availability of oxygen, not decrease it.

• (d) Both low temperature and high atmospheric pressure: This option is incorrect because high atmospheric pressure does not occur at high altitudes. Additionally, while low temperatures can have various physiological effects, they are not the primary reason for low oxygen concentration in the blood and tissues at high altitudes. The main factor is the low atmospheric pressure.

Answer

(a) In pure methanol, molecules are hydrogen bonded. On adding acetone, its molecules get in between the host molecules and break some of the hydrogen bonds between them.

Therefore, the intermolecular attractive forces between the solute-solvent molecules are weaker than those between the solute-solute and solvent-solvent molecules.

On the other hand, other three remaining options will show negative deviation from Raoult’s law where the intermolecular attractive forces between the solute-solvent molecules are stronger than those between the solute-solute and solvent-solvent molecules.

• (b) Chloroform and acetone: In this mixture, chloroform can form hydrogen bonds with acetone. The intermolecular attractive forces between chloroform and acetone molecules are stronger than those between the individual chloroform and acetone molecules, leading to a negative deviation from Raoult’s law.

• (c) Nitric acid and water: Nitric acid and water can form strong hydrogen bonds with each other. The intermolecular attractive forces between nitric acid and water molecules are stronger than those between the individual nitric acid and water molecules, resulting in a negative deviation from Raoult’s law.

• (d) Phenol and aniline: Phenol and aniline can form strong hydrogen bonds with each other. The intermolecular attractive forces between phenol and aniline molecules are stronger than those between the individual phenol and aniline molecules, causing a negative deviation from Raoult’s law.

Answer

(b) Colligative properties depend upon number of solute particles in solution irrespective of their nature. Colligative property is used to determine the molecular mass of particle.

• (a) The nature of the solute particles dissolved in solution: Colligative properties are independent of the chemical nature of the solute particles; they depend solely on the number of solute particles present in the solution.

• (c) The physical properties of the solute particles dissolved in solution: Colligative properties do not depend on the physical properties (such as size, shape, or state) of the solute particles, but rather on their quantity in the solution.

• (d) The nature of solvent particles: Colligative properties are determined by the number of solute particles in the solution and are not influenced by the nature of the solvent particles.

Thinking Process

This process includes concept of van’t Hoff factor and boiling point. Calculate van’t Hoff factor then correlate it with boiling point of solution.

Answer

(b) As we know greater the value of van’t Hoff factor higher will be the elevation in boiling point and hence higher will be the boiling point of solution.

Hence, $1.0 M Na_2 SO_4$ has highest value of boiling point.

• $1.0 M NaOH$: The van’t Hoff factor $(i)$ for NaOH is 2, which is lower than that of Na₂SO₄. Therefore, the elevation in boiling point for NaOH will be less compared to Na₂SO₄.

• $1.0 M NH₄NO₃$: The van’t Hoff factor $(i)$ for NH₄NO₃ is 2, which is lower than that of Na₂SO₄. Consequently, the elevation in boiling point for NH₄NO₃ will be less compared to Na₂SO₄.

• $1.0 M KNO₃$: The van’t Hoff factor $(i)$ for KNO₃ is 2, which is lower than that of Na₂SO₄. As a result, the elevation in boiling point for KNO₃ will be less compared to Na₂SO₄.

Thinking Process

Write the formula of ebullioscopic constant then put the values of their unit and then calculate unit of ebullioscopic constant.

Answer

(a) As we know from elevation in boiling point that

$$ \begin{aligned} \Delta T_{b} & =K_{b} m \\ K_{b} & =\frac{\Delta T_{b}}{m} \\ \text { Unit of } K_{b} & =\frac{\text { unit of } \Delta T_{b}}{\text { unit of } m}=\frac{K}{\text { molality }} \\ & =\frac{K}{mol kg^{-1}}=K mol^{-1} kg \end{aligned} $$

• Option (b): The unit given is mol $kg K^{-1}$ or $K^{-1}$ (molality). This is incorrect because the ebullioscopic constant $K_b$ should have the unit of temperature change per unit molality, not the inverse of temperature or molality.

• Option (c): The unit given is $kg mol^{-1} K^{-1}$ or $K^{-1}$ (molality) ${ }^{-1}$. This is incorrect because it suggests an inverse relationship with temperature and molality, which does not align with the definition of the ebullioscopic constant.

• Option (d): The unit given is $K mol kg^{-1}$ or $K$ (molality). This is incorrect because it implies a direct relationship with molality without considering the inverse nature of molality in the unit of the ebullioscopic constant. The correct unit should be $K$ per molality, which is $K mol^{-1} kg$.

Thinking Process

Calculate value of van’t Hoff factor then correlate it with colligative property of given solution.

Answer

(c) As we know depression in freezing point is directly related to van’t Hoff factor (i) according to which greater the value of $i$ greater will be the depression in freezing point.

Hence, depression in freezing point of glucose is about 3 times of glucose.

• (a) the same: This option is incorrect because the van’t Hoff factor (i) for glucose is 1, while for MgCl₂ it is 3. Since the depression in freezing point is directly proportional to the van’t Hoff factor, the depression in freezing point for MgCl₂ cannot be the same as that for glucose.

• (b) about twice: This option is incorrect because the van’t Hoff factor (i) for MgCl₂ is 3, which means the depression in freezing point for MgCl₂ is three times that of glucose, not twice.

• (d) about six times: This option is incorrect because the van’t Hoff factor (i) for MgCl₂ is 3, which means the depression in freezing point for MgCl₂ is three times that of glucose, not six times.

Answer

(d) When an unripe mango is placed in a concentrated salt solution to prepare pickle then mango loose water due to osmosis and get shrivel.

• (a) It gains water due to osmosis: This is incorrect because in a concentrated salt solution, the water potential outside the mango is lower than inside. Water moves from a region of higher water potential (inside the mango) to a region of lower water potential (the salt solution), causing the mango to lose water, not gain it.

• (b) It loses water due to reverse osmosis: This is incorrect because reverse osmosis involves the movement of water against its concentration gradient, typically requiring an external pressure. In the case of the mango in salt solution, the process is natural osmosis, not reverse osmosis.

• (c) It gains water due to reverse osmosis: This is incorrect because reverse osmosis would require an external pressure to force water into the mango against its concentration gradient. In reality, the mango loses water due to the natural process of osmosis, not reverse osmosis.

Answer

(a) According to definition of osmotic pressure we know that $\pi=C R T$. For concentrated solution $C$ has higher value than dilute solution.

Hence, as concentration of solution increases osmotic pressure will also increase.

• (b) This option is incorrect because, according to the definition of osmotic pressure, $\pi = C R T$. For a dilute solution, the concentration $C$ is lower than that of a concentrated solution. Therefore, the osmotic pressure of a dilute solution will be lower, not higher.

• (c) This option is incorrect because the osmotic pressure is directly proportional to the concentration of the solution. Since a concentrated solution has a higher concentration than a dilute solution, their osmotic pressures cannot be the same.

• (d) This option is incorrect because osmotic pressure can be directly compared between solutions of different concentrations using the formula $\pi = C R T$. Therefore, it is possible to compare the osmotic pressure of a concentrated solution with that of a dilute solution.

Answer

(a) According to definition of depression in freezing point

$$ \Delta T_{f}=K_{f} m $$

where, $K_{f}=$ freezing point depression constant, value of $K_{f}$ depends upon nature of solvent. This is why although the solution have same molality two different solutions of sucrose of same molality prepared in different solvents will have different depression in freezing point.

• (b) The osmotic pressure of a solution is given by the equation $\pi = CRT$ (where $C$ is the molarity of the solution). This statement is actually correct, so it is not a reason for being incorrect.

• (c) The decreasing order of osmotic pressure for $0.01 M$ aqueous solutions of barium chloride, potassium chloride, acetic acid, and sucrose is given as $BaCl_2 > KCl > CH_3COOH > \text{sucrose}$. This statement is correct because osmotic pressure depends on the number of particles in solution (van’t Hoff factor, $i$). $BaCl_2$ dissociates into 3 ions, $KCl$ into 2 ions, $CH_3COOH$ partially dissociates, and sucrose does not dissociate. Therefore, the order is correct.

• (d) According to Raoult’s law, the vapor pressure exerted by a volatile component of a solution is directly proportional to its mole fraction in the solution. This statement is correct, so it is not a reason for being incorrect.

Answer

(b) Number of total ions present in the solution is known as van’t Hoff factors (i).

Hence, correct choice is (b).

• Option (a) is incorrect because the van’t Hoff factor for ( K_2SO_4 ) is 3, not 2.
• Option (c) is incorrect because the van’t Hoff factors for ( KCl ) and ( NaCl ) are 2, not 1, and the van’t Hoff factor for ( K_2SO_4 ) is 3, not 2.
• Option (d) is incorrect because the van’t Hoff factors for ( KCl ) and ( NaCl ) are 2, not 1, and the van’t Hoff factor for ( K_2SO_4 ) is 3, not 1.

Answer

(b) In reverse osmosis, solvent molecules move through a semipermeable membrane from a region of higher concentration of solute to lower concentration.

• (a) This statement is true. Both atmospheric pressure and osmotic pressure are measured in units of pressure, such as Pascals (Pa) or atmospheres (atm).

• (c) This statement is true. The molal depression constant (also known as the cryoscopic constant) depends on the nature of the solvent because it is a property that is specific to each solvent.

• (d) This statement is true. Relative lowering of vapor pressure is a ratio and therefore a dimensionless quantity.

Answer

(a) Value of Henry’s constant $(K_{H})$ increases with increase in temperature representing the decrease in solubility.

• (b) Decreases with increase in temperature: This is incorrect because, according to Henry’s law, the solubility of a gas in a liquid decreases with an increase in temperature, which means Henry’s constant actually increases with temperature.

• (c) Remains constant: This is incorrect because Henry’s constant is temperature-dependent and changes with variations in temperature.

• (d) First increases then decreases: This is incorrect because Henry’s constant consistently increases with an increase in temperature, reflecting the continuous decrease in gas solubility with rising temperature.

Answer

(b) According to Henry’s law

$$ \begin{matrix} & p \propto x \\ \Rightarrow & p=K_{H} x \end{matrix} $$

As value of $K_{H}$ rises solubility of gases decreases.

• (a) Incorrect because Henry’s constant, ( K_H ), is inversely related to the solubility of gases. A higher ( K_H ) value indicates lower solubility, not higher.

• (c) Incorrect because Henry’s constant, ( K_H ), varies for different gases. It is not a universal constant and depends on the nature of the gas and the solvent.

• (d) Incorrect because Henry’s constant, ( K_H ), is directly related to the solubility of gases. It quantifies the relationship between the partial pressure of the gas and its concentration in the solution.

Answer

(b) We know that, if a pressure higher than the osmotic pressure is applied on the solution. the solvent will flow from the solution into the pure solvent through the semi-permeable membrane. This process is called reverse osmosis.

Thus, in this case, water will move from side $(B)$ to side $(A)$ if a pressure greater than osmotic pressure is applied on piston (B).

• Option (a): Water will not move from side (A) to side (B) if a pressure lower than the osmotic pressure is applied on piston (B). For water to move from side (A) to side (B), the pressure applied on piston (B) must be greater than the osmotic pressure, not lower.

• Option (c): Water will not move from side (B) to side (A) if a pressure equal to the osmotic pressure is applied on piston (B). When the pressure applied is equal to the osmotic pressure, there will be no net movement of water across the semi-permeable membrane, as the osmotic pressure is balanced by the applied pressure.

• Option (d): Water will not move from side (A) to side (B) if a pressure equal to the osmotic pressure is applied on piston (A). Similar to option (c), when the pressure applied is equal to the osmotic pressure, there will be no net movement of water across the semi-permeable membrane, as the osmotic pressure is balanced by the applied pressure.

Answer

(b) van’t Hoff factor is the measurement of total number of ions present in the solution. Therefore, greater the concentration of solution greater will be its van’t Hoff factor.

• Option (a) $i_{A}<i_{B}<i_{C}$: This option is incorrect because it suggests that the van’t Hoff factor increases as the concentration decreases. However, the van’t Hoff factor is directly proportional to the concentration of the solution. Therefore, as the concentration decreases, the van’t Hoff factor should also decrease, not increase.

• Option (c) $i_{A}=i_{B}=i_{C}$: This option is incorrect because it assumes that the van’t Hoff factor is the same for all concentrations. In reality, the van’t Hoff factor depends on the concentration of the solution. Since the concentrations of $A$, $B$, and $C$ are different, their van’t Hoff factors cannot be the same.

• Option (d) $i_{A}<i_{B}>i_{C}$: This option is incorrect because it suggests that the van’t Hoff factor for solution $B$ is greater than both $A$ and $C$. This is not possible because the van’t Hoff factor should decrease consistently with the decrease in concentration. Therefore, $i_{B}$ cannot be greater than both $i_{A}$ and $i_{C}$.

Answer

(b) For an ideal solution, the $A-A$ or $B-B$ type intermolecular interaction is near by equal to $A-B$ type interaction. Here, a mixture of bromoethane and chloroethane is an example of ideal solution.

On the other hand chloroform and acetone mixture is an example of non-ideal solution having negative deviation. So, $(A-A)$ or $(B-B)$ interaction must be stronger than $A-B$ interaction. While ethanol-acetone mixture shows positive deviation due to weaker $A-B$ interaction in comparison to $A-A$ or $A-B$ interaction.

• (a) Solution (ii) and (iii) will follow Raoult’s law: This is incorrect because both solutions (ii) and (iii) are examples of non-ideal solutions. Solution (ii) shows positive deviation from Raoult’s law, and solution (iii) shows negative deviation from Raoult’s law. Non-ideal solutions do not follow Raoult’s law.

• (c) Solution (ii) will show negative deviation from Raoult’s law: This is incorrect because solution (ii), which is a mixture of ethanol and acetone, shows positive deviation from Raoult’s law due to weaker $A-B$ interactions compared to $A-A$ or $B-B$ interactions.

• (d) Solution (iii) will show positive deviation from Raoult’s law: This is incorrect because solution (iii), which is a mixture of chloroform and acetone, shows negative deviation from Raoult’s law due to stronger $A-B$ interactions compared to $A-A$ or $B-B$ interactions.

Answer

(a) When salt is added to water to make the solution the vapour pressure of solution get decreases. This is due to decrease in surface covered by solvent molecule which lead to decrease in number of solvent molecule escaping from the surface corresponding to pure solvent.

Hence, vapour pressure also get reduces.

• (b) Vapour pressure in container $(A)$ is less than that in container $(B)$: This is incorrect because adding NaCl to water decreases the vapor pressure of the solution compared to pure water. The presence of solute particles reduces the number of water molecules at the surface, thereby reducing the rate of evaporation and the vapor pressure.

• (c) Vapour pressure is equal in both the containers: This is incorrect because the vapor pressure of a solution (container B) is always lower than that of the pure solvent (container A) due to the presence of solute particles which inhibit the escape of solvent molecules into the vapor phase.

• (d) Vapour pressure in container $(B)$ is twice the vapour pressure in container (A): This is incorrect because the addition of NaCl to water decreases the vapor pressure of the solution. It is not possible for the vapor pressure of the solution to be higher, let alone twice as high, as that of the pure solvent.

Answer

(a) If two liquids $A$ and $B$ form minimum boiling azeotrope at some specific composition then $A-B$ interactions are weaker than those of $A-A$ and $B-B$. It is due to the fact that in case of positive deviation, we get minimum boiling azeotropes whereas in case of negative deviation we get maximum boiling azeotropes.

• (a) $A-B$ interactions are stronger than those between $A-A$ or $B-B$: This is incorrect because for a minimum boiling azeotrope, the $A-B$ interactions are weaker than the $A-A$ and $B-B$ interactions, leading to a positive deviation from Raoult’s law.

• (b) vapour pressure of solution increases because more number of molecules of liquids $A$ and $B$ can escape from the solution: This is incorrect because while it is true that the vapour pressure increases, it is due to the weaker $A-B$ interactions, not necessarily because more molecules can escape.

• (c) vapour pressure of solution decreases because less number of molecules of only one of the liquids escape from the solution: This is incorrect because in the case of a minimum boiling azeotrope, the vapour pressure of the solution actually increases due to the weaker $A-B$ interactions, not decreases.

Thinking Process

To calculate the strength of solution when it is diluted by adding solvent. Write all the given values $M_1, V_1, M_2$ and $V_2$. Then calculate required parameter using formula, $M_1 V_1=M_2 V_2$

where, $\quad V_1=$ volume of solution before dilution

$V_2=$ volume of solution after dilution

$M_1=$ strength of solution before dilution

$M_2=$ strength of solution after dilution

Answer

(d) Given, $M_1=0.02 M, V_1=4 L, M_2=$ ?, $V_2=4 L+1 L=5 L$

As we know,

$$ \begin{aligned} M_1 V_1 & =M_2 V_2 \\ 0.02 \times 4 L & =M_2 \times 5 L \\ M_2 & =\frac{0.08}{5}=0.016 M \end{aligned} $$

• Option (a) 0.004: This option is incorrect because it represents a molality that is too low. The dilution calculation shows that the molality of the resultant solution is higher than 0.004 M.

• Option (b) 0.008: This option is incorrect because it represents a molality that is still too low. The correct molality after dilution is 0.016 M, which is double this value.

• Option (c) 0.012: This option is incorrect because it represents a molality that is slightly lower than the correct value. The correct molality after dilution is 0.016 M, not 0.012 M.

Answer

(a) At specific composition methanol- acetone mixture will form minimum boiling azeotrope and will show positive deviation. This is due to weaker $A-B$ interaction than $A-A$ or $B-B$ interaction. i.e., $A-B<A-A$ and $B-B$

• (b) At specific composition methanol-acetone mixture will form maximum boiling azeotrope and will show positive deviation from Raoult’s law:

  • Incorrect because a maximum boiling azeotrope is formed when the mixture shows negative deviation from Raoult’s law, not positive deviation. Positive deviation indicates weaker interactions between different components, leading to a minimum boiling azeotrope.

• (c) At specific composition methanol-acetone mixture will form minimum boiling azeotrope and will show negative deviation from Raoult’s law:

  • Incorrect because a minimum boiling azeotrope is formed when the mixture shows positive deviation from Raoult’s law, not negative deviation. Negative deviation indicates stronger interactions between different components.

• (d) At specific composition methanol-acetone mixture will form maximum boiling azeotrope and will show negative deviation from Raoult’s law:

  • Incorrect because the given information indicates that the interaction between methanol and acetone is weaker than the interactions within methanol or acetone themselves. This leads to positive deviation from Raoult’s law and the formation of a minimum boiling azeotrope, not a maximum boiling azeotrope.

Thinking Process

Higher the value of $K_{H}$ at a given pressure, the lower is the solubility of the gas in the liquid.

Answer

(c) Value of $K_{H}$ depends upon nature of gases dissolved in water.

Hence, correct order is $Ar<CO_2<CH_4<HCHO$ and correct choice is (c).

• Option (a) is incorrect because it suggests that $HCHO$ is less soluble than $CH_4$, $CO_2$, and $Ar$. However, $HCHO$ has the lowest $K_H$ value, indicating it is the most soluble among the given gases.

• Option (b) is incorrect because it places $CO_2$ as less soluble than $CH_4$. However, $CO_2$ has a higher $K_H$ value than $CH_4$, indicating that $CO_2$ is less soluble than $CH_4$.

• Option (d) is incorrect because it places $CH_4$ as less soluble than $CO_2$. However, $CH_4$ has a lower $K_H$ value than $CO_2$, indicating that $CH_4$ is more soluble than $CO_2$.

Answer

( $a, b$ )

Solubility of gaseous solute in the fixed volume of liquid solvent always depends upon nature of solute but it depends upon pressure at constant temperature and depends upon temperature at constant pressure.

Hence, (a) and (b) both are correct.

• Option (c) is incorrect because it does not consider the nature of the solute, which is a crucial factor affecting solubility.
• Option (d) is incorrect because it only considers pressure and ignores both the nature of the solute and temperature, both of which also affect solubility.

Answer

(c, $d)$

The solution which follows Raoult’s law is known as ideal solution. For an ideal solution intermolecular forces between two benzene molecules are nearly of same strength as those between two toluene molecules. For an ideal solution

$$ \Delta V_{\text {mix }}=0 \text { and } \Delta H_{\text {mix }}=0 $$

Thus, the mixture of benzene and toluene is an example of ideal solution. Option (c) is incorrect as minimum boiling azeotropes are formed by non-ideal solution.

• Option (a) is incorrect because for an ideal solution, the enthalpy of mixing, (\Delta_{\text{mix}} H), is zero. Since benzene and toluene form an ideal solution, (\Delta_{\text{mix}} H = 0).

• Option (b) is incorrect because for an ideal solution, the volume of mixing, (\Delta_{\text{mix}} V), is zero. Since benzene and toluene form an ideal solution, (\Delta_{\text{mix}} V = 0).

Answer

( $a, b$ )

Relative lowering of vapour pressure is a colligative property because

(i) It does not depend upon nature of solute.

(ii) It depends upon number of solute particles.

(iii) It depends upon concentration of non-electrolyte solution.

Hence, (a) and (b) are correct.

• Option (c) is incorrect because it states that the relative lowering of vapour pressure depends on the nature of the solute molecules, which contradicts the definition of a colligative property. Colligative properties depend only on the number of solute particles, not their nature.

• Option (d) is incorrect because it implies that the relative lowering of vapour pressure depends on the nature of solute molecules. As a colligative property, it should depend only on the number of solute particles and not on their specific nature.

Answer

$(a, c)$

van’t Hoff factor (i) is a measure of extent of association or dissociation of solute particles which can be calculated as

$$ \begin{aligned} i & =\frac{\text { normal molar mass }}{\text { abnormal molar mass }} \\ & =\frac{\text { observed colligative property }}{\text { calculated colligative property }} \end{aligned} $$

• Option (b) is incorrect because the van’t Hoff factor (i) is defined as the ratio of the normal molar mass to the abnormal molar mass, not the other way around. Therefore, the correct expression is (i = \frac{\text{normal molar mass}}{\text{abnormal molar mass}}).

• Option (d) is incorrect because the van’t Hoff factor (i) is defined as the ratio of the observed colligative property to the calculated colligative property, not the other way around. Therefore, the correct expression is (i = \frac{\text{observed colligative property}}{\text{calculated colligative property}}).

Answer

(c, $d)$

Isotonic solutions have same osmotic pressure and same concentration. Elevation in boiling point and depression in freezing point are the colligative properties. These two colligative properties depend upon concentration.

As the molar concentration is same for isotonic solutions, so elevation in boiling point and depression in freezing point of isotonic solutions must be same.

• (a) solute: Isotonic solutions do not necessarily have the same solute. They can have different solutes as long as the overall osmotic pressure and concentration are the same.

• (b) density: Isotonic solutions do not need to have the same density. Density is a physical property that depends on both the mass and volume of the solution, and it can vary even if the solutions have the same osmotic pressure and concentration.

Answer

$(b, c)$

Mixtures having same composition in liquid and vapour phase are known as azeotropes. Azeotropes boils at same temperature.

Here, water-nitric acid and water-ethanol mixtures are non-ideal solution. Hence, water-nitric acid and water-ethanol are examples of azeotropes.

While benzene-toluene and $n$-hexane $-n$-heptane are examples of ideal solution.

• Benzene-toluene: This mixture forms an ideal solution, meaning it follows Raoult’s law closely and does not exhibit azeotropic behavior. Therefore, the composition of the liquid and vapor phases will not be the same.

• $n$-hexane-$n$-heptane: This mixture also forms an ideal solution, adhering to Raoult’s law. As a result, it does not form an azeotrope, and the liquid and vapor phases will have different compositions.

Answer

$(b, c)$

The two solutions having same osmotic pressure are known as isotonic solutions. The solute and solvent particles may or may not be same but osmotic pressure must be same.

• Option (a) solute and solvent both are same: This option is incorrect because isotonic solutions are defined by having the same osmotic pressure, not necessarily the same solute and solvent. The solute and solvent can be different as long as the osmotic pressure is the same.

• Option (d) solute is always same solvent may be different: This option is incorrect because isotonic solutions do not require the solute to be the same. The key criterion for isotonic solutions is that they have the same osmotic pressure, regardless of whether the solute or solvent is the same or different.

Answer

$(a, d)$

Depending on the vapour pressures of the pure components 1 and 2, total vapour pressure over the solution decreases or increases with the increase of the mole fraction of component 1.

• Option (b): This option shows a non-linear relationship between total vapor pressure and composition, which is not characteristic of an ideal solution. In an ideal solution, the total vapor pressure should vary linearly with the mole fraction of the components according to Raoult’s Law.

• Option (c): This option depicts a curve that suggests a maximum or minimum in the total vapor pressure, indicating a deviation from ideal behavior. Ideal solutions do not exhibit such extrema; their total vapor pressure changes linearly with composition.

• Option (e): This option shows a constant total vapor pressure regardless of the composition of the solution. This is incorrect for an ideal solution, as the total vapor pressure should vary with the mole fractions of the components according to Raoult’s Law.

Answer

$(a, b)$

When any of one component of binary mixture either solvent or solute is volatile it causes deviation from ideal behaviour and vapour pressure of solution which causes change in colligative property.

Hence, (a) and (b) are correct.

• (c) a gas is dissolved in non-volatile liquid: Colligative properties are typically observed in solutions where the solute is non-volatile. When a gas is dissolved in a non-volatile liquid, the gas can escape from the solution, and this does not significantly affect the colligative properties such as boiling point elevation or freezing point depression.

• (d) a volatile liquid is dissolved in another volatile liquid: When both components of the solution are volatile, the solution does not exhibit colligative properties in the same way as when one component is non-volatile. The presence of two volatile components leads to a more complex interaction and does not result in the straightforward colligative property changes seen with non-volatile solutes.

Answer

Both the components are appearing in the distillate and composition of liquid and vapour is same. This shows that liquids have formed azeotropic mixture and boils at constant temperature hence cannot be separated at this stage by distillation or fractional distillation. Solution having azeotropic nature show large positive or negative deviation from Raoult’s law depending upon intermolecular interaction.

Answer

$NaCl$ is a non-volatile solute. So, its addition to water lowers the vapour pressure of the water. Hence, boiling point of water (solution) increases. Whereas methyl alcohol is more volatile than water.

So, its addition to water increases the total vapour pressure over the solution. It results in the decrease of boiling point.

Answer

If the intermolecular interactions are similar in both constituents, i.e., solute and solvent then solute dissolves in the solvent. e.g., polar solutes dissolve in polar solvents and non-polar solutes in non-polar solvents.

Thus, the statement ’like dissolves like’ proves to be true. e.g., organic compounds dissolve in non-polar organic solvent while polar inorganic compounds (salts) dissolve in polar solvent (water).

Thinking Process

To solve this problem notice the role of temperature in component of concentration term such as volume, mass, number of moles etc.

Answer

Molarity is defined as the number of moles of solute dissolved per litre of a solution. Since, volume depends on temperature and changes with change in temperature, the molarity will also change with change in temperature.

On the other hand, mass does not change with change in temperature, so other concentration terms given in the question also do not do so. Thus, temperature has no effect on the mass but it has significant effect on volume.

Answer

According to Henry’s law $p \propto x \Rightarrow p=K_{H} x$ Higher the value of Henry’s law constant $K_{H}$, the lower is the solubility of the gas in a liquid. Thus, the solubility of a gas in the given liquid can be increased by increasing pressure.

Answer

Aquatic species are more comfortable in cold water due to the presence of more oxygen. Solubility of oxygen in water increases with decrease in temperature as solubility of a gas in given liquid decreases with increase in temperature.

Answer

(a) (i) Henry’s law represents a relation between solubility of gases in liquid and pressure. Scuba drivers when comes towards surface, the pressure gradually decreases. This reduce pressure releases the dissolve gas present in blood and leads to formation of bubbles of nitrogen in the blood.

This creates a painful condition by blocking capillaries known as blends.

(ii) At high altitude atmospheric pressure is low as compared to surface which causes difficulty in breathing. On that condition we feel weakness and discomfort.

(b) Soda water bottle kept at room temperature fizzes on opening due to different pressure inside and outside the bottle. When the bottle is opened to air, the partial pressure of $CO_2$ above the solution decreases. As a result, solubility decreases and hence $CO_2$ bubbles out.

Answer

In pure liquid, the entire surface of liquid is occupied by the molecules of water. When a non-volatile solute, e.g., glucose is dissolved in water, the fraction of surface covered by the solvent molecules gets reduced because some positions are occupied by glucose molecules.

So, number of solvent molecules escaping from the surface is reduced. That is why vapour pressure of aqueous solution of glucose is reduced.

Answer

When salt is spread over snow covered roads, snow starts melting from the surface because depression of freezing point of water takes place due to addition of salt. It helps in clearing of roads.

Hence, the phenomena is depression in freezing point which helps in clearing the snow covered roads in hilly areas.

Answer

Continuous sheets or films (natural or synthetic) which contain a network of submicroscopic holes or pores through which small solvent molecules (water etc.) can pass, but solute molecules of bigger size cannot pass are called semipermeable membrane. e.g., cellophane membrane.

Answer

Since pressure required for the reverse osmosis is very high, so, a suitable material is used for making semipermeable membrane. It is generally cellulose acetate placed over suitable support.

Answer

A. $\rightarrow(4)$

B. $\rightarrow(3)$

C. $\rightarrow(1)$

D. $\rightarrow(2)$

E. $\rightarrow(6)$

F. $\rightarrow(5)$

A. Saturated solution A solution which contains maximum amounts of solute that can be dissolved in a given amount of solvent at a given temperature.

B. Binary solution A solution with two components is known as binary solution.

C. Isotonic solution A solution having same osmotic pressure at a given temperature as that of given solution is known as isotonic solution.

D. Hypotonic solution A solution whose osmotic pressure is less than another is known as hypotonic solution.

E. Solid solution A solution in solid phase is known as solid solution.

F. Hypertonic solution A solution whose osmotic pressure is greater than that of another is known as hypertonic solution.

Answer

A. $\rightarrow(5)$

B. $\rightarrow$ (3)

C. $\rightarrow(4)$

D. $\rightarrow(2)$

E. $\rightarrow(1)$

A. Soda water A solution of gas in liquid. e.g., $CO_2$ in soft drinks.

B. Sugar solution A solution of solid in liquid in which sugar particles (soild) are dissolved in water (liquid).

C. German silver German silver is an alloy which is a solid solution of solid in solid. It is an alloy of $Cu, Zn$ and $Ni$.

D. Air A solution of gas in gas. Air is a mixture of various gases.

E. Hydrogen gas in palladium is an example of solution of gas in solid. This is used as an reducing agent.

Answer

A. $\rightarrow(3)$

B. $\rightarrow$ (5)

C. $\rightarrow(1)$

D. $\rightarrow(1)$

E. $\rightarrow(2)$

A. Raoult’s law Mathematical representation of Raoult’s law

B. Henry’s law $p=K_{H} \cdot x$

$$ p=x_1 p_1^{0}+x_2 p_2^{0} $$

C. Elevation of boiling point Mathematical representation, $\Delta T_{b}=K_{b} \cdot m$

D. Depression in freezing point Mathematical representation, $\Delta T_{f}=K_{f} \cdot m$

E. Osmotic pressure Mathematical representation, $\pi=C R T$.

Answer

A. $\rightarrow$ (4) B. $\rightarrow$ (3) C. $\rightarrow$ (2) D. $\rightarrow$ (5) E. $\rightarrow$ (1)

Answer

(a) Assertion and reason both are correct statements and reason is the correct explanation of assertion.

Volume of solutions is a function of temperature which varies with temperature. Hence, molarity of solution in liquid state changes with temperature.

$$ \text { Molarity }=\frac{\text { moles of solute }}{\text { volume of solution in litre }} $$

Answer

(d) Assertion is wrong statement but reason is correct statement.

When methyl alcohol is added to water, boiling point of water decreases because when a volatile solute is added to a volatile solvent elevation in boiling point is observed.

Answer

(a) Assertion and reason both are correct and reason is correct explanation of assertion. When $NaCl$ is added to water a depression in freezing point is observed. This is due to lowering of vapour pressure of a solution. Lowering of vapour pressure is observed due to intermolecular interaction of solvent-solute particles.

Answer

(b) Assertion and reason both are correct statements but reason is not the correct explanation of assertion.

When a solution is separated from the pure solvent by a semipermeable membrane, the solvent molecules pass through it from pure solvent side to the solution side. Solvent molecules always flow from lower concentration to higher concentration of solution.

Answer

(a) $w / w$ (mass percentage) Mass percentage of a component of a solution can be expressed as

$\text { Mass } $% $\text { of component }=\frac{\text { mass of component in the solution }}{\text { total mass of solution }} \times 100$

Thus, the percentage by mass means the mass of the solute in grams present in $100 g$ of the solution.

(b) $V / V$ (volume percentage) is defined as

$$ \text { Volume percentage }=\frac{\text { volume of the component }}{\text { total volume of solution }} \times 100 $$

Thus, volume percentage means the volume of the liquid solute in $cm^{3}$ present in $100 cm^{3}$ of the solution.

(c) $\boldsymbol{w} / \mathbf{V}$ (mass by volume percentage) = mass of solute dissolved in $100 mL$ of solution.

(d) ppm (parts per million) This parametre is used to express the concentration of very dilute solution.

$$ ppm=\frac{\text { number of parts of component }}{\begin{matrix} \text { total number of parts of all component } \\ \text { of solution } \end{matrix} } \times 10^{6} $$

(e) $\chi$ (mole fraction) Mole fraction is an unitless quantity used to determine extent of any particular component present in total solution.

$$ \chi=\frac{\text { number of moles of the component }}{\text { total number of moles of all components }} $$

(f) $\boldsymbol{M}$ (molarity) Number of moles of solute dissolved in per litre of solution is known as molarity.

$$ M=\frac{\text { number of moles of solute }}{\text { volume of solution in litre }} $$

(g) $m$ (Molality) Molality of any solution can be defined as number of moles of solute dissolved in per $kg$ of solvent.

$$ m=\frac{\text { number of moles of solute }}{\text { mass of solvent in } kg} $$

Answer

According to Raoult’s law for any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.

$$ p_1=p_1^{\circ} x_1 $$

(a) $CHCl_3(l)$ and $CH_2 Cl_2(l)$ both are volatile components.

Hence, for a binary solution in which both components are volatile liquids, the total pressure will be $\quad p=p_1+p_2=x_1 p_1^{\circ}+x_2 p_2^{\circ}$

$$ =x_1 p_1^{\circ}+(1-x_1) p_2^{\circ}=(p_1^{\circ}-p_2^{\circ}) x_1+p_2^{\circ} $$

where, $\quad p=$ total vapour pressure

$p_1=$ partial vapour pressure of component 1

$p_2=$ partial vapour pressure of component 2

(b) $NaCl(s)$ and $H_2 O(l)$ both are non-volatile components.

Hence, for a solution containing non-volatile solute, the Raoult’s law is applicable only to vaporisable component and total vapour pressure can be written as

$$ p=p_1=x_1 p_1^{\circ} $$

Answer

The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. For an ideal solution $\Delta V_{\text {mix }}=O$ and $\Delta V_{\text {mix }}=O$. The ideal behaviour of the solutions can be explained by considering two components $A$ and $B$.

In pure components, the intermolecular attractive interactions will be of $A-A$ type and $B-B$ type, whereas in the binary solutions in addition to these two, $A-B$ type of interaction will also be present. If $A-A$ and $B-B$ intermolecular forces are nearly equal to those between $A-B$, this leads to the formation of ideal solution e.g., solution of $n$-hexane and $n$-heptane.

When a solution does not obey-Raoult’s law over the entire range of concentration, then it is called non-ideal solution. The vapour pressure of such a solution is either higher or lower, than that predicted by Raoult’s law.

If it is higher, the solution exhibits positive deviation and if it is lower it exhibits negative deviation from Raoult’s law. In case of positive deviation, $A-B$ interactions are weaker than those between $A-A$ or $B-B$. i.e., the attractive forces between solute solvent molecules are weaker than those between solute-solute and solvent-solvent molecules e.g., mixture of ethanol and acetone.

For such solutions

$$ \Delta H_{\text {mixing }}=+ve \text { and } \Delta V_{\text {mixing }}=+ve $$

On the other hand, in case of negative deviation the intermolecular attractive forces between $A-A$ and $B-B$ are weaker than those between $A-B$ molecules. Thus, the escaping tendency of $A$ and $B$ types of molecules from the solution becomes less than from the pure liquids i.e., mixture of chloroform and acetone.

For such solution

$\Delta H_{\text {mix }}=- \text { ve and } \Delta V_{\text {mix }}=-v e$

Answer

The solution or mixture having same composition in liquid as well as in vapour phase and boils at a constant temperature is known as azeotropes. Due to constant composition it can’t be separated by fractional distillation. There are two types of azeotropes

(i) Minimum boiling azeotropes Solutions which show large positive deviation from Raoult’s law form minimum boiling azeotropes at a specific composition. e.g., ethanol —water mixture

(ii) Maximum boiling azeotropes Solutions which show large negative deviation from Raoult’s law form maximum boiling azeotropes. e.g., solution having composition $68 %$ $HNO_3$ and $32 %$ water by mass.

Answer

This phenomenon is called endo osmosis, i.e., movement of water inside the raisin and shown with the help of diagram as

The process of osmosis is of immense biological as well as industrial important. It is evident from the following examples.

(i) Movement of water from soil into plant roots and subsequently into upper portion of the plant is partly due to osmosis.

(ii) Preservation of meat against bacterial action by addition of salt.

(iii) Preservation of fruits against bacterial action by adding sugar. Bacterium in canned fruit loses water through the process of osmosis and become inactive.

(iv) Reverse-osmosis is used in desalination of water.

Answer

(i) In animals, water moves into different parts of the body under the effect of the process of osmosis.

(ii) Stretching of leaves, flower, etc., is also controlled by osmosis.

(iii) Osmosis helps in rapid growth of the plants and germination of seeds.

(iv) Different movements of plants such as opening and closing of flowers etc, are controlled by osmosis.

Thinking Process

The question can be answered using the concept of solubility, osmosis, reverse-osmosis, hypertonic solution and hypotonic solution.

Answer

When egg is placed in mineral acid solution outershell of egg dissolves.

Egg is now removed and placed in hypertonic solution. Size of egg get reduced and egg shrivels due to osmosis. Egg is now placed in a bottle with narrow neck. Finally on adding hypotonic solution egg regain its shape due to osmosis.

Diagramatically it can be represented as

Answer

Certain compounds when dissolved in suitable solvents either dissociate or associate. e.g., ethanoic acid dimerises in benzene due to $H$-bonding, while in water, it dissociates and forms ions.

As a result the number of chemical species in solution increases or decreases as compared to the number of chemical species of solute added to form the solution.

Since, the magnitude of colligative property depends on the number of solute particles, it is expected that the molar mass determined on the basis of colligative properties will either higher or lower than the expected value or the normal value and is called abnormal molar mass.

In order to account for the extent of dissociation or association of molecules in solution, van’t Hoff introduced a factor, $i$, known as the van’t Hoff factor.

$$ \begin{aligned} & i= \frac{\text { expected molar mass }}{\text { abnormal molar mass }}=\frac{\text { observed colligative property }}{\text { calculated colligative property }} \\ & \text { total number of moles of particles after } \\ &= \frac{\text { association or dissociation }}{\text { total number of moles of particles }} \\ & \text { before association or dissociation } \end{aligned} $$