3. Comprehensive Theory Notes
3.1 Expression of Concentration of Solutions
The concentration of a solution is a measure of the amount of solute dissolved in a specific amount of solvent or solution. Understanding concentration is critical for preparing standard solutions in the laboratory.
Effect of Temperature: It is highly crucial to remember for CBSE boards that concentration terms involving volume (like Molarity, Volume percentage) are temperature-dependent because liquids expand or contract with temperature. Conversely, concentration terms involving only mass (like Molality, Mole fraction, Mass percentage) are temperature-independent.
3.2 Solubility
Solubility is the maximum amount of solute that can be dissolved in a specified amount of solvent at a specific temperature. It depends on the nature of the solute, nature of the solvent, temperature, and pressure ("Like dissolves like").
Solubility of a Solid in a Liquid
- Dissolution: When a solid solute is added to the solvent, some solute dissolves, and its concentration increases in the solution.
- Crystallization: Some solute particles in solution collide with the solid solute particles and get separated out of the solution.
- Dynamic Equilibrium: A stage is reached when the rate of dissolution equals the rate of crystallization. The solution at this point is called a saturated solution.
- Effect of Temperature: According to Le Chatelier's Principle, if the dissolution process is endothermic ($\Delta H_{sol} > 0$), solubility increases with a rise in temperature. If the process is exothermic ($\Delta H_{sol} < 0$), solubility decreases with a rise in temperature.
- Effect of Pressure: Pressure does not have any significant effect on the solubility of solids in liquids because they are highly incompressible.
Solubility of a Gas in a Liquid (Henry's Law)
Unlike solids, gases are highly compressible. Thus, pressure has a profound effect on the solubility of a gas in a liquid. This relationship is quantified by Henry's Law.
Henry's Law Statement: The partial pressure of the gas in the vapour phase ($p$) is directly proportional to the mole fraction of the gas ($x$) in the solution.
$$ p = K_H \times x $$
Where $K_H$ is Henry's Law constant.
Important properties of $K_H$:
- The value of $K_H$ depends on the nature of the gas.
- Higher the value of $K_H$ at a given pressure, the lower is the solubility of the gas in the liquid.
- The value of $K_H$ increases with an increase in temperature. This means that gas solubility decreases as temperature rises. This is why aquatic species are more comfortable in cold water rather than in warm water (more dissolved oxygen in cold water).
Applications of Henry's Law:
- In carbonated beverages: To increase the solubility of $CO_2$ in soft drinks and soda water, the bottles are sealed under high pressure.
- In deep-sea diving: Scuba divers breathe air at high pressure underwater. This increases the solubility of atmospheric gases ($N_2$, $O_2$) in blood. When divers ascend quickly to the surface, the pressure decreases rapidly, releasing dissolved nitrogen in the form of bubbles in the blood. This blocks capillaries and creates a painful, dangerous medical condition called "bends". To avoid this, scuba tanks are filled with air diluted with helium ($11.7\% \text{ He}, 56.2\% \text{ N}_2, 32.1\% \text{ O}_2$). Helium is less soluble in blood.
- At high altitudes: The partial pressure of oxygen at high altitudes is less than at ground level. This leads to a low concentration of oxygen in the blood and tissues of mountain climbers, causing weakness and loss of clear thinking, a condition known as anoxia.
3.3 Vapour Pressure and Raoult's Law
Vapour pressure is the pressure exerted by the vapour in equilibrium with its liquid phase at a constant temperature.
Raoult's Law for Liquid-Liquid Solutions (Volatile solute)
Raoult's Law Statement: For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in the solution.
$$ p_A = p_A^0 x_A \quad \text{and} \quad p_B = p_B^0 x_B $$
Using Dalton's law of partial pressures, the total pressure of the solution is: $$ P_{total} = p_A + p_B = p_A^0 x_A + p_B^0 x_B = p_A^0 + (p_B^0 - p_A^0)x_B $$
Raoult's Law for Solid-Liquid Solutions (Non-volatile solute)
When a non-volatile solute (e.g., salt, sugar) is added to a volatile solvent (e.g., water), the vapour pressure of the solution decreases. This happens because the non-volatile solute particles occupy some fraction of the surface area of the liquid, reducing the escape of solvent molecules into the vapour phase. In this case, Raoult's law simplifies to stating that the vapour pressure of the solution is directly proportional to the mole fraction of the solvent alone.
3.4 Ideal and Non-Ideal Solutions
| Property |
Ideal Solutions |
Non-Ideal (Positive Deviation) |
Non-Ideal (Negative Deviation) |
| Raoult's Law | Obeys at all concentrations: $P = p_A^0x_A + p_B^0x_B$ | Does not obey: $P > p_A^0x_A + p_B^0x_B$ | Does not obey: $P < p_A^0x_A + p_B^0x_B$ |
| Interactions | A-B interactions = A-A and B-B interactions | A-B interactions < A-A and B-B interactions | A-B interactions > A-A and B-B interactions |
| Enthalpy Change ($\Delta H_{mix}$) | $\Delta H_{mix} = 0$ (No heat absorbed/evolved) | $\Delta H_{mix} > 0$ (Endothermic, cooling occurs) | $\Delta H_{mix} < 0$ (Exothermic, heating occurs) |
| Volume Change ($\Delta V_{mix}$) | $\Delta V_{mix} = 0$ | $\Delta V_{mix} > 0$ (Volume expands) | $\Delta V_{mix} < 0$ (Volume contracts) |
| Examples | n-Hexane + n-Heptane, Benzene + Toluene, Bromoethane + Chloroethane | Ethanol + Acetone, Carbon disulfide + Acetone, Ethanol + Water | Phenol + Aniline, Chloroform + Acetone, Nitric acid + Water |
Deep Dive into Deviations: Why does a mixture of Ethanol and Acetone show a positive deviation? Pure ethanol molecules are strongly hydrogen-bonded. When acetone is added, its molecules get in between the ethanol molecules, breaking the strong hydrogen bonds. This weakens the intermolecular interactions, making it easier for molecules to escape into the vapour phase, thus increasing the vapour pressure beyond Raoult's law predictions.
Why does a mixture of Chloroform and Acetone show a negative deviation? A mixture of chloroform and acetone forms a new intermolecular hydrogen bond between the oxygen atom of acetone and the hydrogen atom of chloroform. This makes escaping to the vapour phase more difficult, resulting in a lowered vapour pressure.
3.5 Azeotropes (Constant Boiling Mixtures)
Azeotropes are binary mixtures having the same composition in liquid and vapour phase and boil at a constant temperature. Because their vapour composition is exactly the same as their liquid composition, their components cannot be separated by fractional distillation.
- Minimum boiling azeotrope: Formed by solutions showing large positive deviations from Raoult's law. The azeotrope boils at a temperature lower than either of the pure components. Example: 95.4% Ethanol + 4.6% Water by volume.
- Maximum boiling azeotrope: Formed by solutions showing large negative deviations from Raoult's law. The azeotrope boils at a temperature higher than either of the pure components. Example: 68% Nitric acid + 32% Water by mass.
3.6 Colligative Properties
Properties of dilute solutions containing a non-volatile solute that depend only on the number of solute particles (molecules or ions) present in the solution, and not on the chemical nature of the solute, are called Colligative Properties. There are four colligative properties:
1. Relative Lowering of Vapour Pressure (RLVP)
When a non-volatile solute is added to a solvent, the vapour pressure of the solution is lower than that of the pure solvent. The ratio of the lowering of vapour pressure ($\Delta p = p_A^0 - p_A$) to the vapour pressure of the pure solvent ($p_A^0$) is called the relative lowering of vapour pressure. It is equal to the mole fraction of the solute.
$$ \frac{p_A^0 - p_A}{p_A^0} = x_B = \frac{n_B}{n_A + n_B} $$
2. Elevation in Boiling Point ($\Delta T_b$)
The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to atmospheric pressure. Since adding a non-volatile solute lowers the vapour pressure, the solution must be heated to a higher temperature to make its vapour pressure equal to 1 atm. This increase is called the elevation in boiling point.
$$ \Delta T_b = K_b \times m $$ Where $K_b$ is the Ebullioscopic Constant or Molal Elevation Constant (Units: $K \text{ kg mol}^{-1}$). It is the elevation in boiling point when 1 mole of solute is dissolved in 1 kg of solvent.
3. Depression in Freezing Point ($\Delta T_f$)
The freezing point is the temperature at which the liquid and solid phases of a substance are in dynamic equilibrium and have the same vapour pressure. Since a solution has a lower vapour pressure than the pure solvent, its freezing point will be lower.
$$ \Delta T_f = K_f \times m $$ Where $K_f$ is the Cryoscopic Constant or Molal Depression Constant. A prime real-life application of this property is the use of Ethylene Glycol as an anti-freeze in car radiators during freezing winter conditions, as it drastically lowers the freezing point of water.
4. Osmosis and Osmotic Pressure ($\pi$)
Osmosis: The spontaneous 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 semi-permeable membrane (SPM).
Osmotic Pressure: The minimum excess pressure that must be applied to the solution side to prevent the entry of solvent into the solution through the SPM.
$$ \pi = C \times R \times T = \left(\frac{n}{V}\right)RT $$
Advantage of Osmotic Pressure: It is widely used to determine the molar masses of proteins, polymers, and other macromolecules because osmotic pressure is measured at room temperature, and its magnitude is large even for very dilute solutions.
- Isotonic Solutions: Two solutions having the same osmotic pressure at a given temperature. No osmosis occurs between them. (e.g., 0.9% w/v NaCl solution is isotonic with human red blood cells).
- Hypertonic Solution: A solution with higher osmotic pressure. If RBCs are placed in a hypertonic solution (>0.9% NaCl), water flows out of the cells, causing them to shrink (plasmolysis).
- Hypotonic Solution: A solution with lower osmotic pressure. If RBCs are placed in a hypotonic solution (<0.9% NaCl), water flows into the cells, causing them to swell and potentially burst.
Reverse Osmosis (RO): If a pressure larger than the osmotic pressure is applied to the solution side, the direction of osmosis is reversed. Solvent molecules move from the solution to the pure solvent side. This principle is heavily utilized in the desalination of seawater to obtain drinking water.
3.7 Abnormal Molar Mass and Van't Hoff Factor ($i$)
The molar mass calculated using colligative properties is sometimes different from the actual/theoretical molar mass. This happens due to two reasons:
- Dissociation: Molecules of electrolytes (acids, bases, salts like NaCl, KCl) break into multiple ions when dissolved in water. E.g., one molecule of NaCl gives two particles ($Na^+$ and $Cl^-$). Thus, the number of particles increases, colligative property increases, and the calculated molar mass decreases.
- Association: Molecules of some substances join together to form dimers, trimers, etc., due to hydrogen bonding. E.g., Acetic acid ($CH_3COOH$) forms dimers in benzene. The number of particles decreases, colligative property decreases, and calculated molar mass increases.
To correct this discrepancy, Jacobus Henricus van 't Hoff introduced the Van't Hoff factor ($i$).
$$ i = \frac{\text{Normal molar mass}}{\text{Abnormal (observed) molar mass}} $$
$$ i = \frac{\text{Total number of moles of particles after association/dissociation}}{\text{Number of moles of particles before association/dissociation}} $$
- For dissociation, $i > 1$.
- For association, $i < 1$.
- For a non-electrolyte (like glucose, urea, sucrose), $i = 1$.
4. Most Important CBSE Board Questions
The questions below are curated strictly based on the latest CBSE pattern, including Assertion-Reasoning, Case-Based Questions, and high-frequency numericals.
Section A: 1 Mark Questions (MCQs & A-R)
Q1. Which of the following is dependent on temperature? 1 Mark
- Molality
- Molarity
- Mole fraction
- Mass percentage
Answer: (b) Molarity
Explanation: Molarity involves the volume of the solution. Since volume changes with temperature due to thermal expansion or contraction of the liquid, molarity is temperature-dependent. The others depend only on mass, which is constant.
Q2. A solution of two liquids boils at a temperature higher than the boiling point of either of them. The solution is: 1 Mark
- an ideal solution
- a non-ideal solution showing positive deviation
- a non-ideal solution showing negative deviation
- a minimum boiling azeotrope
Answer: (c) a non-ideal solution showing negative deviation
Explanation: A mixture that boils at a higher temperature means its vapour pressure is abnormally low. A lower vapour pressure than Raoult's law expectation means it exhibits a negative deviation, forming a maximum boiling azeotrope.
Q3. For an ideal solution, the correct option is: 1 Mark
- $\Delta_{mix} H = 0$ at constant T, P
- $\Delta_{mix} V \neq 0$ at constant T, P
- $\Delta_{mix} S = 0$ at constant T, P
- $\Delta_{mix} G = 0$ at constant T, P
Answer: (a) $\Delta_{mix} H = 0$ at constant T, P
Explanation: For an ideal solution, there is no change in enthalpy ($\Delta_{mix}H = 0$) or volume ($\Delta_{mix}V = 0$) upon mixing. Entropy of mixing ($\Delta_{mix}S$) is always positive, and Gibbs free energy ($\Delta_{mix}G$) is always negative for any spontaneous mixing process.
Q4. The van't Hoff factor ($i$) for a dilute aqueous solution of the strong electrolyte barium chloride is: 1 Mark
- 1
- 2
- 3
- 0
Answer: (c) 3
Explanation: Barium chloride ($BaCl_2$) dissociates completely in dilute solution into $Ba^{2+}$ and $2 Cl^-$ ions. Total ions produced = 1 + 2 = 3. Since it's a strong electrolyte, $\alpha = 1$, thus $i = 3$.
Q5. The value of Henry's constant $K_H$ is: 1 Mark
- greater for gases with higher solubility.
- greater for gases with lower solubility.
- constant for all gases.
- not related to the solubility of gases.
Answer: (b) greater for gases with lower solubility.
Explanation: According to Henry's Law, $p = K_H \times x \implies x = p / K_H$. Thus, solubility ($x$) is inversely proportional to $K_H$ at a constant pressure.
Assertion-Reasoning Question Directions:
(A) Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
(B) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
(C) Assertion is correct but Reason is incorrect.
(D) Assertion is incorrect but Reason is correct.
Q6. Assertion (A): Aquatic species are more comfortable in cold water rather than in warm water.
Reason (R): Different gases have different $K_H$ values at the same temperature.
1 Mark
Answer: (B)
Explanation: The assertion is true because the solubility of oxygen is higher in cold water (solubility of gases in liquids increases with a decrease in temperature). The reason is also a factual statement. However, the correct explanation for the assertion should be that $K_H$ for oxygen increases with a rise in temperature (lowering solubility), not that different gases have different $K_H$ values.
Q7. Assertion (A): When NaCl is added to water, a depression in the freezing point is observed.
Reason (R): The lowering of vapour pressure of a solution causes depression in the freezing point.
1 Mark
Answer: (A)
Explanation: Adding a non-volatile solute (NaCl) to water lowers the vapour pressure of the solution. Because of this lowered vapour pressure, the solution will now freeze at a lower temperature than the pure solvent. Thus, both statements are true and the reason correctly explains the assertion.
Section B: 2 Mark Questions (Very Short Answer)
Q8. State Raoult's law for a solution containing volatile components. How does Raoult's law become a special case of Henry's law? 2 Marks
Answer:
Raoult's Law: It states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.
$p_A = p_A^0 x_A$
Special Case: According to Henry's law, the partial pressure of a gas in a liquid is $p = K_H x$. According to Raoult's law, the partial vapour pressure of a volatile component is $p = p^0 x$. When we compare both, they are structurally identical. The only difference is the proportionality constant. Raoult's law becomes a special case of Henry's law when $K_H$ becomes equal to $p^0$ (vapour pressure of pure component).
Q9. What are azeotropes? Give one example of a minimum boiling azeotrope. 2 Marks
Answer:
Azeotropes: These are liquid mixtures that boil at a constant temperature without undergoing any change in their composition. Their liquid and vapour phases have the exact same mole fractions. They cannot be separated by fractional distillation.
Example: A mixture of 95.4% ethanol and 4.6% water (by volume) forms a minimum boiling azeotrope. This occurs because the mixture shows a large positive deviation from Raoult's law.
Q10. Calculate the molarity of a solution containing 5g of NaOH in 450 mL of solution. 2 Marks
Answer:
Given: Mass of solute ($NaOH$), $W_B = 5\text{ g}$.
Molar mass of $NaOH$ ($M_B$) = $23 (Na) + 16 (O) + 1 (H) = 40\text{ g mol}^{-1}$.
Volume of solution, $V = 450\text{ mL}$.
Formula: $M = \frac{W_B \times 1000}{M_B \times V(\text{in mL})}$
Calculation:
$M = \frac{5 \times 1000}{40 \times 450}$
$M = \frac{5000}{18000} = \frac{5}{18} = 0.278\text{ M}$
The molarity of the solution is 0.278 mol/L.
Section C: 3 Mark Questions (Short Answer)
Q11. Define osmotic pressure. Why is osmotic pressure considered a colligative property? A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15K. 3 Marks
Answer:
Osmotic Pressure: It is the excess pressure that must be applied to the solution side to prevent the flow of pure solvent into the solution through a semi-permeable membrane.
It is considered a colligative property because its value depends only on the number of solute particles (molar concentration of the solution) and not on the chemical nature of the solute particles ($\pi = CRT$).
Numerical Solution:
Molar mass of cane sugar ($C_{12}H_{22}O_{11}$) = $342\text{ g mol}^{-1}$
Molar mass of glucose ($C_6H_{12}O_6$) = $180\text{ g mol}^{-1}$
Step 1: Find $K_f$ using the cane sugar data.
5% by mass of cane sugar means 5g sugar in 95g water.
Molality ($m_1$) = $\frac{W_B \times 1000}{M_B \times W_A} = \frac{5 \times 1000}{342 \times 95} = 0.1539\text{ mol/kg}$
$\Delta T_f$ for cane sugar = $273.15\text{ K} - 271\text{ K} = 2.15\text{ K}$
$\Delta T_f = K_f \times m_1 \implies 2.15 = K_f \times 0.1539 \implies K_f = \frac{2.15}{0.1539} = 13.97\text{ K kg mol}^{-1}$
Step 2: Apply to glucose solution.
5% by mass of glucose means 5g glucose in 95g water.
Molality ($m_2$) = $\frac{W_B \times 1000}{M_B \times W_A} = \frac{5 \times 1000}{180 \times 95} = 0.2924\text{ mol/kg}$
$\Delta T_f (\text{glucose}) = K_f \times m_2 = 13.97 \times 0.2924 = 4.085\text{ K}$
Freezing point of glucose solution = $T_f^0 - \Delta T_f = 273.15 - 4.085 = \mathbf{269.065\text{ K}}$.
Q12. What is meant by positive and negative deviations from Raoult's law and how is the sign of $\Delta_{mix} H$ related to positive and negative deviations from Raoult's law? 3 Marks
Answer:
Positive Deviation: When the vapour pressure of a solution is higher than that predicted by Raoult's law, it is a positive deviation. This happens when the attractive forces between solute-solvent (A-B) molecules are weaker than those between pure solvent-solvent (A-A) and pure solute-solute (B-B) molecules. The escaping tendency of molecules increases.
Relation to $\Delta_{mix} H$: Because old stronger bonds break and new weaker bonds form, energy is required. The process is endothermic, thus $\Delta_{mix} H > 0$ (Positive).
Negative Deviation: When the vapour pressure of a solution is lower than that predicted by Raoult's law. This happens when the attractive forces between solute-solvent (A-B) molecules are stronger than those between pure molecules (A-A and B-B). The escaping tendency decreases.
Relation to $\Delta_{mix} H$: Because stronger bonds are formed, energy is released. The process is exothermic, thus $\Delta_{mix} H < 0$ (Negative).
Q13. Boiling point of water at 750 mm Hg is 99.63 °C. How much sucrose is to be added to 500 g of water such that it boils at 100 °C? [$K_b$ for water = 0.52 K kg mol⁻¹] 3 Marks
Answer:
Given:
Mass of water (solvent), $W_A = 500\text{ g}$
Boiling point of pure water at given pressure, $T_b^0 = 99.63^\circ\text{C}$
Boiling point of solution, $T_b = 100^\circ\text{C}$
$K_b = 0.52\text{ K kg mol}^{-1}$
Molar mass of sucrose ($C_{12}H_{22}O_{11}$), $M_B = 342\text{ g mol}^{-1}$
Step 1: Calculate elevation in boiling point ($\Delta T_b$)
$\Delta T_b = T_b - T_b^0 = 100 - 99.63 = 0.37^\circ\text{C}$ (or $0.37\text{ K}$, since a temperature difference in Celsius is equal to Kelvin).
Step 2: Apply the formula
$\Delta T_b = \frac{K_b \times W_B \times 1000}{M_B \times W_A}$
$0.37 = \frac{0.52 \times W_B \times 1000}{342 \times 500}$
$0.37 = \frac{0.52 \times W_B \times 2}{342}$
$W_B = \frac{0.37 \times 342}{1.04}$
$W_B = \mathbf{121.67\text{ g}}$
Hence, 121.67 g of sucrose must be added.
Section D: 4 Mark Case-Based Questions
Q14. Read the passage given below and answer the following questions:
Osmosis is a phenomenon of great importance in biological systems. A biological membrane is semi-permeable, allowing water to pass through but restricting the passage of larger molecules and ions. The pressure required to just stop the osmotic flow of solvent is called osmotic pressure. In intravenous (IV) injections, the concentration of the fluid injected must match the osmotic pressure of blood plasma. A 0.9% (mass/volume) sodium chloride solution is isotonic with blood plasma. If red blood cells are placed in a solution containing more than 0.9% sodium chloride, water will flow out of the cells, causing them to shrink. Conversely, if placed in a solution with less than 0.9% sodium chloride, water flows into the cells, causing them to swell and burst. People consuming a lot of salt experience water retention in tissue cells and intercellular spaces because of osmosis. The resulting puffiness is called edema.
(i) What is meant by isotonic solutions? (1 Mark)
(ii) Why is a 0.9% (w/v) NaCl solution used in intravenous injections? (1 Mark)
(iii) (a) A patient suffering from high blood pressure is advised to take less salt. Why? (2 Marks)
OR
(iii) (b) Calculate the osmotic pressure of a 0.1 M aqueous solution of NaCl at 300K, assuming complete dissociation. (R = 0.082 L atm K⁻¹ mol⁻¹) (2 Marks)
Answers:
(i) Solutions having the exact same osmotic pressure at a given temperature are called isotonic solutions. When separated by a semipermeable membrane, no net osmosis occurs between them.
(ii) A 0.9% (w/v) NaCl solution is exactly isotonic with human blood plasma. Injecting this ensures that the red blood cells neither shrink (plasmolysis) nor swell and burst, thus maintaining cellular integrity.
(iii) (a) (Internal Choice 1): High salt intake increases the concentration of Na+ ions in the blood plasma and extracellular fluids. Due to osmosis, water moves from tissues into the blood vessels to dilute the salt, increasing the blood volume. This places extra pressure on blood vessel walls, raising blood pressure. Additionally, it causes water retention in intercellular spaces, leading to edema.
(iii) (b) (Internal Choice 2):
Given: $C = 0.1\text{ M}$, $T = 300\text{ K}$, $R = 0.082\text{ L atm K}^{-1}\text{mol}^{-1}$.
Since NaCl is a strong electrolyte and dissociates completely ($NaCl \rightarrow Na^+ + Cl^-$), the van't Hoff factor $i = 2$.
Osmotic Pressure ($\pi$) = $i \times C \times R \times T$
$\pi = 2 \times 0.1 \times 0.082 \times 300$
$\pi = 0.2 \times 24.6$
$\pi = \mathbf{4.92\text{ atm}}$.
Q15. Read the passage given below and answer the following questions:
When a non-volatile solute is added to a solvent, the vapour pressure of the solvent decreases. This phenomenon is directly responsible for other colligative properties like elevation of boiling point and depression of freezing point. The extent of this lowering depends strictly on the number of solute particles. However, if the solute associates or dissociates in the solution, the observed colligative properties deviate from the calculated values. Van't Hoff introduced a factor '$i$' to account for the extent of association or dissociation.
(i) State the relationship between the relative lowering of vapour pressure and the mole fraction of the solute. (1 Mark)
(ii) What will be the van't Hoff factor for $K_2SO_4$ if it is 100% dissociated? (1 Mark)
(iii) (a) Out of 0.1 M urea, 0.1 M NaCl, and 0.1 M $CaCl_2$, which will have the highest boiling point and why? (2 Marks)
OR
(iii) (b) Explain why molecular mass of acetic acid calculated by observing the colligative property in benzene is found to be roughly double its actual molecular mass. (2 Marks)
Answers:
(i) The relative lowering of vapour pressure of a solution containing a non-volatile solute is exactly equal to the mole fraction of the solute in the solution. $(p^0 - p) / p^0 = x_{solute}$.
(ii) $K_2SO_4$ dissociates into $2 K^+$ and $1 SO_4^{2-}$ ion. Total particles = 3. Since it is 100% dissociated ($\alpha=1$), the van't Hoff factor $i = 3$.
(iii) (a) (Internal Choice 1): Elevation in boiling point ($\Delta T_b$) is a colligative property and is directly proportional to the total number of particles (value of $i$).
For Urea (non-electrolyte): $i = 1$
For NaCl: $i = 2$ ($Na^+, Cl^-$)
For $CaCl_2$: $i = 3$ ($Ca^{2+}, 2Cl^-$)
Since $CaCl_2$ produces the maximum number of particles in solution, it will have the maximum elevation in boiling point, and thus the highest boiling point.
(iii) (b) (Internal Choice 2): Acetic acid ($CH_3COOH$) undergoes dimerization in benzene due to intermolecular hydrogen bonding. Two molecules of acetic acid join to form one single entity (dimer). Therefore, the number of particles is reduced to half. Since colligative properties are inversely proportional to molar mass, reducing the particles by half results in the observed colligative property being half the expected value, and the calculated molar mass appears to be double the normal molecular mass (120 g/mol instead of 60 g/mol).
Section E: 5 Mark Long Answer Questions
Q16. (a) Define Henry's Law. Mention any two of its applications. (2 Marks)
(b) The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase. (3 Marks)
5 Marks
Answer:
(a) Henry's Law: The partial pressure of a gas in vapour phase ($p$) is directly proportional to the mole fraction of the gas ($x$) in the solution. $p = K_H \cdot x$.
Applications:
1. To increase the solubility of $CO_2$ in soft drinks, bottles are sealed under high pressure.
2. Scuba divers use cylinders filled with air diluted with helium to avoid "bends," a painful condition caused by nitrogen bubbles forming in the blood when the diver ascends and pressure decreases.
(b) Numerical Calculation:
Given:
$p_A^0 = 450\text{ mm Hg}$
$p_B^0 = 700\text{ mm Hg}$
$P_{total} = 600\text{ mm Hg}$
Step 1: Composition of the liquid mixture ($x_A, x_B$)
According to Raoult's Law:
$P_{total} = p_A^0 x_A + p_B^0 x_B$
Since $x_A + x_B = 1$, we can substitute $x_B = 1 - x_A$.
$600 = 450 x_A + 700(1 - x_A)$
$600 = 450 x_A + 700 - 700 x_A$
$600 - 700 = -250 x_A$
$-100 = -250 x_A \implies x_A = \frac{100}{250} = \mathbf{0.4}$
Therefore, $x_B = 1 - 0.4 = \mathbf{0.6}$
Composition of liquid mixture is 0.4 mole fraction of A and 0.6 mole fraction of B.
Step 2: Composition of the vapour phase ($y_A, y_B$)
According to Dalton's Law of partial pressures in the vapour phase:
$p_A = y_A \times P_{total}$
First, calculate partial pressures $p_A$ and $p_B$:
$p_A = p_A^0 x_A = 450 \times 0.4 = 180\text{ mm Hg}$
$p_B = p_B^0 x_B = 700 \times 0.6 = 420\text{ mm Hg}$
Now, find mole fractions in vapour phase:
$y_A = \frac{p_A}{P_{total}} = \frac{180}{600} = \mathbf{0.30}$
$y_B = \frac{p_B}{P_{total}} = \frac{420}{600} = \mathbf{0.70}$
(Alternatively, $y_B = 1 - y_A = 1 - 0.30 = 0.70$)
Q17. (a) What happens when we place blood cells in water (hypotonic solution)? Give reason. (1 Mark)
(b) What type of deviation from Raoult's law is shown by a mixture of ethanol and acetone? Give reason. (1.5 Marks)
(c) Calculate the freezing point of an aqueous solution containing 10.50 g of $MgBr_2$ in 200 g of water, assuming complete dissociation of $MgBr_2$.
(Molar mass of $MgBr_2$ = 184 g/mol, $K_f$ for water = 1.86 K kg/mol) (2.5 Marks)
5 Marks
Answer:
(a) When red blood cells are placed in pure water (a hypotonic solution), the osmotic pressure outside the cell is lower than inside. Through osmosis, water rapidly flows into the blood cells. The cells swell, and because animal cells lack a rigid cell wall, they eventually burst (hemolysis).
(b) A mixture of ethanol and acetone shows a positive deviation from Raoult's law.
Reason: Pure ethanol possesses strong intermolecular hydrogen bonding. When acetone is added, its molecules intercede between ethanol molecules, breaking some of these hydrogen bonds. This weakens the overall intermolecular forces (A-B interactions < A-A / B-B interactions), making it easier for molecules to escape into the vapour phase. This results in an increased vapour pressure.
(c) Numerical Calculation:
Given: Mass of solute ($MgBr_2$), $W_B = 10.50\text{ g}$
Molar mass of solute, $M_B = 184\text{ g mol}^{-1}$
Mass of solvent (water), $W_A = 200\text{ g} = 0.2\text{ kg}$
$K_f = 1.86\text{ K kg mol}^{-1}$
Step 1: Determine the van't Hoff factor ($i$)
$MgBr_2$ dissociates completely (as given): $MgBr_2 \rightarrow Mg^{2+} + 2Br^-$
Total number of ions = $1 + 2 = 3$. Since dissociation is 100%, $i = 3$.
Step 2: Calculate Molality ($m$)
$m = \frac{W_B \times 1000}{M_B \times W_A} = \frac{10.50 \times 1000}{184 \times 200} = \frac{10500}{36800} = 0.2853\text{ m}$
Step 3: Calculate depression in freezing point ($\Delta T_f$)
$\Delta T_f = i \times K_f \times m$
$\Delta T_f = 3 \times 1.86 \times 0.2853$
$\Delta T_f = 5.58 \times 0.2853 = \mathbf{1.59\text{ K}} \text{ (or } 1.59^\circ\text{C})$
Step 4: Find freezing point of solution
Freezing point of pure water $T_f^0 = 0^\circ\text{C}$ (or $273.15\text{ K}$)
$T_f \text{ (solution)} = T_f^0 - \Delta T_f = 0 - 1.59 = \mathbf{-1.59^\circ\text{C}}$ (or $271.56\text{ K}$)
