Chapter 2: Solutions
Exhaustive Revision Guide - 50 Subjective Questions with Detailed Solutions
Click on any question to reveal its answer.
1. What is a solution? Define solute and solvent.
Solution: A solution is a homogeneous mixture of two or more non-reacting substances whose composition can be varied within certain limits.
Solute: The component of the solution which constitutes the smaller part and is dissolved in the solvent is called the solute.
Solvent: The component of the solution which constitutes the larger part and acts as a dissolving medium is called the solvent.
2. Differentiate between saturated, unsaturated, and supersaturated solutions.
- Unsaturated Solution: A solution in which more solute can be dissolved at the same temperature without raising it.
- Saturated Solution: A solution in which no more solute can be dissolved at a specific temperature. It is in dynamic equilibrium with the undissolved solid solute.
- Supersaturated Solution: A metastable solution containing more solute than the saturated solution at that temperature. It usually crystallizes out the excess solute upon scratching or seeding.
3. Define Solubility. State its unit.
Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a specified temperature to form a saturated solution.
Unit: It is generally expressed in $\text{mol L}^{-1}$ or $\text{g L}^{-1}$.
4. What are the factors affecting the solubility of a solid in a liquid?
The solubility of a solid in a liquid depends on:
- Nature of Solute and Solvent: "Like dissolves like." Polar solutes dissolve in polar solvents (e.g., NaCl in water), and non-polar solutes dissolve in non-polar solvents (e.g., iodine in benzene).
- Temperature: If the dissolution is endothermic, solubility increases with a rise in temperature. If exothermic, solubility decreases.
- Pressure: Pressure has no significant effect on the solubility of solids in liquids because solids and liquids are highly incompressible.
5. How does temperature affect the solubility of a gas in a liquid? Give a reason.
The solubility of a gas in a liquid decreases with an increase in temperature.
Reason: Dissolution of a gas in a liquid is an exothermic process (gas + solvent $\rightleftharpoons$ solution + heat). According to Le Chatelier's principle, increasing the temperature shifts the equilibrium backward, favoring the release of the dissolved gas.
6. State Henry's Law and give its mathematical expression.
Henry's Law: It states that the solubility of a gas in a liquid at constant temperature is directly proportional to the pressure of the gas above the solution.
Mathematical Expression: $S \propto P \Rightarrow S = K_H \cdot P$
Where $S$ is the solubility (in mol/L), $P$ is the partial pressure of the gas, and $K_H$ is Henry's law constant.
7. What are the exceptions (limitations) to Henry's law?
Henry's law is applicable only if:
- The gas behaves like an ideal gas (pressure is not too high, temperature is not too low).
- The gas does not undergo any chemical reaction with the solvent. (e.g., $NH_3$ in water reacts to form $NH_4OH$, so it doesn't obey Henry's law well).
- The gas does not dissociate or associate in the solvent.
8. State any two applications of Henry's Law.
- In carbonated beverages: Soft drinks and soda water bottles are sealed under high pressure to increase the solubility of $CO_2$.
- In deep-sea diving: Scuba divers breathe compressed air, leading to high solubility of $N_2$ in blood. To prevent "the bends" (painful bubbles of $N_2$ forming in blood upon surfacing), diving tanks are diluted with Helium, which is less soluble.
9. Why do aquatic animals find it more comfortable to live in cold water rather than warm water?
The solubility of oxygen (a gas) in water decreases as the temperature increases. In cold water, the concentration of dissolved oxygen is higher. Aquatic animals rely on this dissolved oxygen for respiration. Hence, they feel more comfortable in cold water where oxygen is abundant compared to warm water.
10. Define Mole Fraction. Write its formula.
Mole Fraction ($x$): It is defined as the ratio of the number of moles of a particular component to the total number of moles of all the components present in the solution.
For a binary solution containing $n_1$ moles of solvent and $n_2$ moles of solute:
Mole fraction of solvent, $x_1 = \frac{n_1}{n_1 + n_2}$
Mole fraction of solute, $x_2 = \frac{n_2}{n_1 + n_2}$
Note: $x_1 + x_2 = 1$. It is a dimensionless quantity and independent of temperature.
11. Define Vapor Pressure of a liquid.
Vapor Pressure: The pressure exerted by the vapor of a liquid in dynamic equilibrium with its liquid phase at a given constant temperature in a closed container is called the vapor pressure of the liquid.
12. State Raoult's Law for a solution containing two volatile liquid components.
Raoult's Law: It states that for a solution of volatile liquids, the partial vapor pressure of any volatile component in the solution is directly proportional to its mole fraction in the solution.
Mathematically: $P_1 = P_1^\circ \cdot x_1$ and $P_2 = P_2^\circ \cdot x_2$
Where $P_1$ is partial pressure, $P_1^\circ$ is vapor pressure of pure component 1, and $x_1$ is its mole fraction.
13. What are Ideal Solutions? State their characteristics.
Ideal Solutions: Solutions that obey Raoult's law exactly over the entire range of concentrations and temperatures.
Characteristics:
- Follows Raoult's law: $P_{total} = P_1 + P_2$.
- Enthalpy of mixing is zero: $\Delta H_{mix} = 0$ (No heat evolved or absorbed).
- Volume of mixing is zero: $\Delta V_{mix} = 0$ (Total volume is sum of individual volumes).
- Intermolecular interactions between solute-solvent (A-B) are identical to solute-solute (B-B) and solvent-solvent (A-A) interactions.
Example: Benzene + Toluene.
14. Distinguish between ideal and non-ideal solutions.
| Ideal Solution | Non-Ideal Solution |
|---|---|
| Obeys Raoult's law at all concentrations. | Does not obey Raoult's law. |
| $\Delta H_{mix} = 0$ | $\Delta H_{mix} \neq 0$ |
| $\Delta V_{mix} = 0$ | $\Delta V_{mix} \neq 0$ |
| A-B interactions are similar to A-A and B-B. | A-B interactions are stronger or weaker than A-A and B-B. |
15. Explain positive deviation from Raoult's law with an example.
In solutions showing positive deviation, the solute-solvent (A-B) attractive forces are weaker than the pure solvent-solvent (A-A) and solute-solute (B-B) forces.
As a result, molecules escape more easily into the vapor phase, making the total vapor pressure higher than predicted by Raoult's law ($P_{total} > P_1^\circ x_1 + P_2^\circ x_2$). Also, $\Delta H_{mix} > 0$ and $\Delta V_{mix} > 0$.
Example: Ethanol + Acetone.
16. Explain negative deviation from Raoult's law with an example.
In solutions showing negative deviation, the solute-solvent (A-B) attractive forces are stronger than the pure solvent-solvent (A-A) and solute-solute (B-B) forces.
As a result, the escaping tendency of molecules decreases, making the total vapor pressure lower than predicted by Raoult's law ($P_{total} < P_1^\circ x_1 + P_2^\circ x_2$). Also, $\Delta H_{mix} < 0$ and $\Delta V_{mix} < 0$.
Example: Chloroform + Acetone (strong hydrogen bond forms between them).
17. Define Azeotropes. What are minimum and maximum boiling azeotropes?
Azeotropes: Binary mixtures having the same composition in liquid and vapor phase and which boil at a constant temperature are called azeotropes. They cannot be separated by fractional distillation.
- Minimum Boiling Azeotrope: Formed by solutions showing large positive deviation from Raoult's law. They boil at a lower temperature than either pure component. (e.g., 95.5% Ethanol + Water).
- Maximum Boiling Azeotrope: Formed by solutions showing large negative deviation from Raoult's law. They boil at a higher temperature than either pure component. (e.g., 68% Nitric acid + Water).
18. What are Colligative Properties? Name the four colligative properties.
Colligative Properties: The physical properties of solutions that depend entirely on the number of solute particles present in the given amount of solvent, and not on the nature of the solute particles, are called colligative properties.
The four properties are:
- Relative lowering of vapor pressure.
- Elevation in boiling point.
- Depression in freezing point.
- Osmotic pressure.
19. Why does the vapor pressure of a liquid decrease when a non-volatile solute is added?
Vapor pressure depends on the escape of solvent molecules from the surface of the liquid. When a non-volatile solute is added, some of the surface area is occupied by the solute particles, which do not vaporize. This decreases the fraction of the surface covered by solvent molecules, lowering the rate of evaporation and consequently lowering the vapor pressure of the solution.
20. Define lowering of vapor pressure and relative lowering of vapor pressure.
Lowering of Vapor Pressure ($\Delta P$): It is the difference between the vapor pressure of the pure solvent ($P_1^\circ$) and the vapor pressure of the solution ($P_1$).
$\Delta P = P_1^\circ - P_1$
Relative Lowering of Vapor Pressure: It is the ratio of the lowering of vapor pressure to the vapor pressure of the pure solvent.
$\text{RLVP} = \frac{P_1^\circ - P_1}{P_1^\circ}$
21. State Raoult's Law for a solution containing a non-volatile solute.
For a solution of a non-volatile solute, Raoult's law states that the relative lowering of vapor pressure is equal to the mole fraction of the solute in the solution.
Mathematically: $$\frac{P_1^\circ - P_1}{P_1^\circ} = x_2$$
Where $x_2$ is the mole fraction of the non-volatile solute.
22. Derive the formula to determine the molar mass of a solute from the relative lowering of vapor pressure.
From Raoult's law: $\frac{P_1^\circ - P_1}{P_1^\circ} = x_2$
We know, $x_2 = \frac{n_2}{n_1 + n_2}$. For dilute solutions, $n_2 << n_1$, so $n_1 + n_2 \approx n_1$.
Therefore, $x_2 \approx \frac{n_2}{n_1} = \frac{W_2 / M_2}{W_1 / M_1} = \frac{W_2 \cdot M_1}{W_1 \cdot M_2}$
Substituting this in Raoult's law equation:
$$\frac{P_1^\circ - P_1}{P_1^\circ} = \frac{W_2 \cdot M_1}{W_1 \cdot M_2}$$
Where $W_2, M_2$ are mass and molar mass of solute, and $W_1, M_1$ are mass and molar mass of solvent.
23. Define Boiling Point of a liquid.
The boiling point of a liquid is the temperature at which its vapor pressure becomes equal to the external atmospheric pressure.
24. Why is the boiling point of a solution containing a non-volatile solute always higher than that of the pure solvent?
The addition of a non-volatile solute lowers the vapor pressure of the solvent. To make this lowered vapor pressure equal to the external atmospheric pressure and make the solution boil, the solution must be heated to a higher temperature than the pure solvent. This increase in temperature is called the elevation of boiling point ($\Delta T_b$).
25. Define Ebullioscopic constant (Molal elevation constant, $K_b$) and state its unit.
Ebullioscopic Constant ($K_b$): It is defined as the elevation in boiling point when one mole of a non-volatile solute is dissolved in 1 kg (1000 g) of the solvent. It is the elevation in boiling point of a 1 molal solution.
Unit: $\text{K kg mol}^{-1}$.
26. Derive the relationship between molar mass of a solute and elevation of boiling point.
Experimentally, elevation of boiling point ($\Delta T_b$) is directly proportional to molality ($m$).
$\Delta T_b = K_b \cdot m$
We know molality $m = \frac{n_2}{W_1(\text{in kg})} = \frac{W_2 / M_2}{W_1 / 1000} = \frac{1000 \cdot W_2}{M_2 \cdot W_1}$
Substituting $m$ in the equation:
$$\Delta T_b = \frac{1000 \cdot K_b \cdot W_2}{M_2 \cdot W_1}$$
Rearranging for molar mass ($M_2$): $$M_2 = \frac{1000 \cdot K_b \cdot W_2}{\Delta T_b \cdot W_1}$$
27. Define Freezing Point.
The freezing point of a substance is the temperature at which its liquid phase and solid phase are in equilibrium, meaning they have the same vapor pressure.
28. Why does a solution of a non-volatile solute freeze at a lower temperature than the pure solvent?
Adding a non-volatile solute lowers the vapor pressure of the liquid solvent. Therefore, the vapor pressure of the solution will become equal to the vapor pressure of the solid solvent at a much lower temperature compared to the pure solvent. This results in a depression of the freezing point ($\Delta T_f$).
29. Define Cryoscopic constant (Molal depression constant, $K_f$) and state its unit.
Cryoscopic Constant ($K_f$): It is defined as the depression in freezing point produced when one mole of a non-volatile solute is dissolved in 1 kg (1000 g) of the solvent. It is the freezing point depression of a 1 molal solution.
Unit: $\text{K kg mol}^{-1}$.
30. Derive the formula for determining molar mass from depression in freezing point.
Depression of freezing point ($\Delta T_f$) is directly proportional to molality ($m$).
$\Delta T_f = K_f \cdot m$
Molality $m = \frac{1000 \cdot W_2}{M_2 \cdot W_1}$
Substituting $m$ in the equation:
$$\Delta T_f = \frac{1000 \cdot K_f \cdot W_2}{M_2 \cdot W_1}$$
Rearranging for molar mass ($M_2$): $$M_2 = \frac{1000 \cdot K_f \cdot W_2}{\Delta T_f \cdot W_1}$$
31. What is an anti-freeze? Give an example and state its application.
Anti-freeze: A substance added to a solvent (like water) to lower its freezing point significantly, preventing it from freezing in cold climates.
Example: Ethylene glycol is used as an anti-freeze in the radiators of cars.
Application: In cold countries, water in car radiators would freeze and burst the pipes. Adding ethylene glycol depresses the freezing point below $0^\circ \text{C}$, keeping the coolant in liquid form.
32. What is a semipermeable membrane (SPM)? Give two examples.
Semipermeable Membrane: A membrane that allows the passage of solvent molecules (like water) through its microscopic pores but prevents the passage of larger solute molecules.
Examples: Animal bladder, parchment paper, cellophane, synthetic membranes like cellulose acetate, copper ferrocyanide ($Cu_2[Fe(CN)_6]$).
33. Define Osmosis.
Osmosis: The spontaneous net flow of solvent molecules from a region of lower solute concentration (pure solvent or dilute solution) to a region of higher solute concentration (concentrated solution) through a semipermeable membrane.
34. Define Osmotic Pressure ($\pi$).
Osmotic Pressure: The excess hydrostatic pressure that must be applied to the solution side to just stop the inward flow of pure solvent across a semipermeable membrane is called osmotic pressure.
35. State van't Hoff equation for osmotic pressure and show its relation to molar mass.
van't Hoff equation states that osmotic pressure ($\pi$) is proportional to molarity ($C$) and absolute temperature ($T$).
$$\pi = C \cdot R \cdot T$$
Where $C = \frac{n_2}{V} = \frac{W_2}{M_2 \cdot V}$ ($R$ is the gas constant).
Substituting $C$:
$$\pi = \frac{W_2 \cdot R \cdot T}{M_2 \cdot V} \Rightarrow M_2 = \frac{W_2 \cdot R \cdot T}{\pi \cdot V}$$
36. Define Isotonic, Hypertonic, and Hypotonic solutions.
- Isotonic Solutions: Two solutions having the same osmotic pressure at a given temperature. No net osmosis occurs between them.
- Hypertonic Solution: A solution with a higher osmotic pressure (higher concentration) relative to another solution.
- Hypotonic Solution: A solution with a lower osmotic pressure (lower concentration) relative to another solution.
37. What happens when red blood cells (RBCs) are placed in a 0.5% NaCl solution? Give a reason.
The fluid inside RBCs is isotonic with 0.9% (w/v) NaCl solution. A 0.5% NaCl solution is hypotonic compared to the RBC fluid.
When placed in a 0.5% NaCl solution, water will flow into the cell via endosmosis. As a result, the RBCs will swell and may even burst.
38. What is Reverse Osmosis (RO)? State its most important application.
Reverse Osmosis: If a pressure larger than the osmotic pressure is applied to the solution side, pure solvent flows out of the solution across the semipermeable membrane into the pure solvent side. This process reverses the natural direction of osmosis.
Application: Desalination of seawater. When high pressure is applied to seawater, pure water is squeezed through a cellulose acetate membrane, leaving salts behind, providing drinking water.
39. Why is osmotic pressure considered the best colligative property for determining the molar mass of macromolecules like proteins and polymers?
Osmotic pressure is preferred because:
- It is measured at room temperature, preventing the degradation of temperature-sensitive biomolecules like proteins.
- The magnitude of osmotic pressure is measurable even for very dilute solutions containing macromolecules with huge molar masses, whereas $\Delta T_b$ or $\Delta T_f$ values would be too small to measure accurately.
- It uses molarity instead of molality, which is easier to prepare for polymeric solutions.
40. What is meant by abnormal molar mass? Why does it happen?
When the molar mass calculated using colligative properties differs from the theoretical (true) molar mass, it is called an abnormal molar mass.
Reason: This occurs when the solute undergoes either dissociation (e.g., $NaCl \rightarrow Na^+ + Cl^-$) or association (e.g., acetic acid dimerizing in benzene) in the solution. This changes the actual number of particles in the solution, changing the colligative property value and thus giving an incorrect molar mass.
41. Define van't Hoff factor (i).
van't Hoff factor ($i$): It is defined as the ratio of the observed (experimental) colligative property to the theoretical (calculated) colligative property.
$$i = \frac{\text{Observed colligative property}}{\text{Theoretical colligative property}}$$
Since colligative property $\propto \frac{1}{\text{Molar Mass}}$, it can also be written as:
$$i = \frac{\text{Theoretical Molar Mass}}{\text{Observed Molar Mass}}$$
42. Write the modified equations for all four colligative properties incorporating the van't Hoff factor.
To correct for association or dissociation, the theoretical formulas are multiplied by '$i$':
- RLVP: $\frac{P_1^\circ - P_1}{P_1^\circ} = i \cdot x_2$ (for dilute solutions, $i \cdot \frac{n_2}{n_1}$)
- Elevation in B.P: $\Delta T_b = i \cdot K_b \cdot m$
- Depression in F.P: $\Delta T_f = i \cdot K_f \cdot m$
- Osmotic Pressure: $\pi = i \cdot C \cdot R \cdot T$
43. Relate the van't Hoff factor ($i$) with the degree of dissociation ($\alpha$).
For a solute that dissociates into $n$ ions:
$$\alpha = \frac{i - 1}{n - 1}$$
Where $\alpha$ is the degree of dissociation, $i$ is the van't Hoff factor, and $n$ is the number of moles of ions produced per mole of solute formula unit.
44. What is the value of the van't Hoff factor ($i$) for $K_2SO_4$ assuming complete dissociation?
$K_2SO_4$ dissociates as follows:
$$K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$$
One mole of $K_2SO_4$ yields $2 + 1 = 3$ moles of ions.
For complete dissociation ($\alpha = 1$), the van't Hoff factor $i = n$. Therefore, $i = 3$.
45. Explain why the van't Hoff factor for acetic acid dissolved in benzene is less than 1.
In a non-polar solvent like benzene, acetic acid molecules undergo association (dimerization) due to intermolecular hydrogen bonding.
$$2 CH_3COOH \rightleftharpoons (CH_3COOH)_2$$
Two molecules associate to act as a single particle. This decreases the total number of particles in the solution. Since $i = \frac{\text{Observed particles}}{\text{Theoretical particles}}$, and observed particles are fewer, $i < 1$ (it is approximately 0.5 for complete dimerization).
46. Define Molality ($m$). Why is it preferred over molarity for studying colligative properties involving temperature changes?
Molality ($m$): The number of moles of solute dissolved per kilogram (1000g) of the solvent.
Preference: Molarity involves the volume of the solution, which changes with temperature due to expansion or contraction. Molality involves the mass of the solvent, which is independent of temperature. Thus, molality is temperature-independent, making it ideal for boiling point and freezing point experiments.
47. What is Ebullioscopy and Cryoscopy?
- Ebullioscopy: The method of determining the molar mass of a non-volatile solute by measuring the elevation in the boiling point of the solvent.
- Cryoscopy: The method of determining the molar mass of a non-volatile solute by measuring the depression in the freezing point of the solvent.
48. Draw and explain the vapor pressure vs temperature graph showing elevation of boiling point.
In a graph of Vapor Pressure (y-axis) vs Temperature (x-axis):
- There are two upward-curving lines: one for the pure solvent (upper curve) and one for the solution (lower curve, since VP is lowered).
- A horizontal line is drawn at 1 atm (external atmospheric pressure).
- The temperature where the pure solvent curve intersects 1 atm is $T_b^\circ$ (boiling point of pure solvent).
- The temperature where the solution curve intersects 1 atm is $T_b$ (boiling point of solution).
- Because the solution curve is lower, it intersects the 1 atm line at a higher temperature. Therefore, $T_b > T_b^\circ$, showing $\Delta T_b = T_b - T_b^\circ$.
49. Which will have a higher boiling point: 0.1 M NaCl or 0.1 M Glucose? Give reason.
0.1 M NaCl will have a higher boiling point.
Reason: Elevation in boiling point is a colligative property that depends on the number of particles. Glucose is a non-electrolyte and does not dissociate ($i=1$), providing 0.1 moles of particles. NaCl is a strong electrolyte and dissociates completely into $Na^+$ and $Cl^-$ ions ($i=2$), providing $0.1 \times 2 = 0.2$ moles of particles. More particles mean greater elevation in boiling point.
50. Calculate the mass percent of benzene in a solution containing 30g of benzene dissolved in 120g of carbon tetrachloride.
Mass of solute (benzene) = $30$ g.
Mass of solvent (carbon tetrachloride) = $120$ g.
Total mass of solution = $30 + 120 = 150$ g.
$$\text{Mass percent} = \frac{\text{Mass of component}}{\text{Total mass of solution}} \times 100$$
$$\text{Mass percent of benzene} = \frac{30}{150} \times 100 = 20\%$$
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