CBSE Class 12 Chemistry Solutions Assignment Set 03

Very Short Answer Type Questions

Question. Differentiate between molarity and molality of a solution.
Answer: Molarity is the quantity of solute moles contained in one liter of solution, and it changes with temperature. Molality represents the moles of solute dissolved in one kilogram of the solvent, and it remains unaffected by temperature variations.
In simple words: Molarity measures moles of solute per liter of liquid, which shifts when temperature changes. Molality measures moles of solute per kilogram of solvent, which stays the same regardless of heat.

Exam Tip: Keep in mind that molarity is volume-dependent and thus changes with temperature, whereas molality is mass-dependent and remains constant.

Question. What type of semiconductor is obtained when silicon is doped with arsenic?
Answer: An n-type semiconductor is produced when silicon is doped with arsenic.
In simple words: Doping silicon with arsenic gives extra electrons, making a negative-type (n-type) semiconductor.

Exam Tip: Remember that doping Group 14 elements (like Silicon) with Group 15 elements (like Arsenic) always yields n-type semiconductors due to the presence of extra free electrons.

Question. What is meant by ‘reverse osmosis’?
Answer: When a pressure greater than the osmotic pressure is exerted on the solution, the solvent flows from the solution side to the pure solvent side through a semipermeable membrane. This phenomenon is termed reverse osmosis (R.O.).
In simple words: If you apply high pressure to a salty solution, clean water is forced backward through a special filter, leaving the salt behind.

Exam Tip: Define reverse osmosis clearly by stating that the applied pressure must exceed the osmotic pressure, forcing the solvent to flow in the direction opposite to normal osmosis.

Question. What are isotonic solutions?
Answer: Isotonic solutions are those that exhibit the identical osmotic pressure under the same conditions. For instance, a 0.9% mass/volume sodium chloride solution is isotonic with human blood cells.
In simple words: Isotonic solutions have the same strength or osmotic pressure as each other, meaning liquid won't flow between them.

Exam Tip: Always mention the example of 0.9% mass/volume NaCl being isotonic with human blood cells to secure full marks.

Question. Some liquids on mixing form ‘azeotropes’. What are ‘azeotropes’?
Answer: An azeotrope is a liquid mixture that possesses a constant composition and boils at a constant temperature just like a pure liquid, without any change in its ratio.
In simple words: Azeotropes are special liquid mixtures that boil at a single temperature and cannot be separated by boiling because their vapor has the exact same mix.

Exam Tip: Define an azeotrope as a constant boiling mixture that distills without any change in its composition.

Question. What type of intermolecular attractive interaction exists in the pair of methanol and acetone?
Answer: Dipole-dipole attractions occur between the solute and solvent molecules in a mixture of methanol and acetone.
In simple words: Methanol and acetone molecules attract each other using dipole-dipole forces.

Exam Tip: Identify both compounds as polar molecules to correctly deduce that they interact via solute-solvent dipole-dipole forces.

Question. Out of BaCl2 and KCl, which one is more effective in causing coagulation of a negatively charged colloidal Sol? Give reason.
Answer: Barium chloride (\( \text{BaCl}_2 \)) is more efficient in bringing about coagulation because the barium ion (\( \text{Ba}^{2+} \)) carries a higher positive charge than the potassium ion (\( \text{K}^+ \)). According to Hardy-Schulze rule, greater valence of coagulating ion causes faster coagulation.
In simple words: Barium chloride is better because its barium ions have a double positive charge, which neutralizes negative particles much faster than potassium's single charge.

Exam Tip: Always cite the Hardy-Schulze rule and compare the charges of the active coagulating ions (\( \text{Ba}^{2+} \) versus \( \text{K}^+ \)) explicitly.

Short Answer Type Questions-I (SA-I)

Question. Differentiate between molality and molarity of a solution. What is the effect of change in temperature of a solution on its molality and molarity?
Answer: Molarity (\( M \)) is the quantity of solute moles contained in one liter of solution: \[ M = \frac{w \times 1000}{\text{molar mass} \times V} \] Molality (\( m \)) refers to the number of solute moles dissolved in one kilogram of solvent: \[ m = \frac{w \times 1000}{M_2 \times W} \] The mathematical correlation linking molarity to molality is expressed as: \[ m = \frac{M}{d - \frac{M \cdot M_2}{1000}} \] Molarity is temperature-dependent since liquid volume alters with thermal changes. Conversely, molality is independent of temperature because it is based on mass, which remains constant.
In simple words: Molarity uses the volume of the solution, which changes when heated or cooled. Molality uses the weight of the solvent, which never changes with temperature.

Exam Tip: For temperature effects, emphasize that volume depends on temperature while mass does not, explaining why only molarity is temperature-dependent.

Question. Non-ideal solutions exhibit either positive or negative deviations from Raoult’s law. What are these deviations and why are they caused? Explain with one example for each type.
Answer: Negative Deviation: In these mixtures, the partial vapour pressure of every constituent, as well as the total vapour pressure, is lower than expected from Raoult's law. This occurs because the solute-solvent (\( A-B \)) attractive forces are stronger than the individual (\( A-A \)) and (\( B-B \)) forces. Example: A mix of chloroform (\( \text{CHCl}_3 \)) and acetone.
Positive Deviation: In these mixtures, the actual vapour pressure of the system is higher than the value predicted by Raoult's law. This happens because the (\( A-B \)) attractive interactions are weaker than the (\( A-A \)) and (\( B-B \)) interactions. Example: A mixture of acetone and benzene.
In simple words: Negative deviation happens when different molecules love each other more than their own kind, lowering vapor pressure. Positive deviation happens when they dislike each other, making them escape into vapor more easily.

Exam Tip: Explain deviations in terms of the relative strength of intermolecular forces, specifically comparing A-B interactions with A-A and B-B interactions.

Question. Define the terms, ‘osmosis’ and ‘osmotic pressure’. What is the advantage of using osmotic pressure as compared to other colligative properties for the determination of molar masses of solutes in solutions?
Answer: Osmosis: The spontaneous passage of solvent particles from a pure solvent or dilute solution into a concentrated solution across a semipermeable membrane is known as osmosis.
Osmotic pressure: The minimum external pressure that must be applied to the solution to halt the flow of solvent molecules through the semipermeable membrane is called osmotic pressure.
The prime benefit of using this technique is that measurements are taken at room temperature and use molarity instead of molality.
In simple words: Osmosis is solvent flowing naturally into a thicker solution. Osmotic pressure is the squeeze needed to stop it. This method is great because it works at normal room temperature and is easy to measure for large molecules like proteins.

Exam Tip: When listing advantages, highlight that osmotic pressure can be measured at room temperature, which is crucial for biomolecules like proteins that decompose at high temperatures.

Question. A 1.00 molal aqueous solution of trichloroacetic acid (CCl3COOH) is heated to its boiling point. The solution has the boiling point of 100.18°C. Determine the van’t Hoff factor for trichloroacetic acid. (Kb for water = 0.512 K kg mol^–1)
Answer: We know that: \[ \Delta T_b = i \cdot K_b \cdot m \] Given: Boiling point of solution = \( 100.18^\circ\text{C} \) Boiling point of pure water = \( 100^\circ\text{C} \) \( \Delta T_b = 100.18 - 100 = 0.18^\circ\text{C} = 0.18\text{ K} \) Molality, \( m = 1.00\text{ m} \) \( K_b = 0.512\text{ K kg mol}^{-1} \) Substituting these values: \( 0.18 = i \times 0.512 \times 1.00 \)
\( \implies i = \frac{0.18}{0.512} \approx 0.35 \) Thus, the van't Hoff factor is \( 0.35 \).
In simple words: We find the boiling point elevation, which is 0.18 degrees. Using the formula with the constant and concentration, we divide to get a van't Hoff factor of 0.35.

Exam Tip: Be sure to calculate \( \Delta T_b \) first by subtracting the boiling point of pure water (\( 100^\circ\text{C} \)) from the solution's boiling point, and remember that \( i \) is a dimensionless factor.

Question. Define the following terms : (i) Mole fraction (ii) Isotonic solutions (iii) van’t Hoff factor (iv) Ideal solution
Answer: (i) Mole fraction: This is the ratio of the moles of a single component to the total moles of all components present in the mixture.
(ii) Isotonic solutions: These are solutions that exhibit identical osmotic pressure at the same temperature.
(iii) van't Hoff factor: This represents the ratio of the normal molecular mass to the observed abnormal molecular mass, given by: \[ i = \frac{\text{normal molar mass}}{\text{abnormal molar mass}} \]
(iv) Ideal solution: A solution that strictly follows Raoult's law across the entire range of concentrations is termed an ideal solution.
In simple words: Mole fraction is your share of the total moles. Isotonic means having the same pressure. The van't Hoff factor compares expected mass to actual mass. An ideal solution obeys the basic law perfectly.

Exam Tip: Ensure that each of the four definitions is written clearly, stating any underlying conditions such as "at a given temperature" or "over the entire range of concentration."

Question. Explain why aquatic species are more comfortable in cold water rather than in warm water.
Answer: Aquatic creatures require dissolved oxygen to breathe. Since the solubility of gases in liquids drops as the temperature rises, less dissolved oxygen is present in warm water during summer. Consequently, marine organisms are much more comfortable in colder water where oxygen levels are higher.
In simple words: Fish need oxygen dissolved in the water. Cold water holds more oxygen than warm water, making it much easier for them to breathe in the winter.

Exam Tip: Connect the temperature change directly to the solubility of gases using Henry's Law to demonstrate a deeper thermodynamic understanding.

Question. State Raoult’s law. How is it formulated for solutions of non-volatile solutes?
Answer: Raoult's law states that for a solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction. For a solution containing a non-volatile solute, the vapour pressure of the solution (\( p_1 \)) is directly proportional to the mole fraction of the solvent (\( X_1 \)). Mathematically: \[ p_1 = p_1^0 \cdot X_1 \] Where \( p_1^0 \) is the vapour pressure of the pure solvent.
In simple words: Raoult's law says that the pressure of a solution depends on the amount of solvent in it. When you add solid solute that doesn't evaporate, the pressure equals the pure solvent's pressure multiplied by the solvent's mole fraction.

Exam Tip: Clarify that for non-volatile solutes, the solute has zero vapour pressure, so the total vapour pressure of the solution is entirely due to the solvent.

Question. State Henry's law and mention two of its important applications.
Answer: Henry's law: This law states that the partial pressure of a gas in the vapour phase is directly proportional to the mole fraction of that gas dissolved in the solution.
Applications of Henry’s law:
(i) To boost the solubility of carbon dioxide in soft drinks and soda, packaging is done under extremely high pressure.
(ii) To prevent a painful condition known as bends, deep-sea divers utilize air tanks diluted with helium, which has low solubility.
In simple words: Henry's law says that pushing a gas harder forces more of it to dissolve in a liquid. This is why soda is bottled under high pressure and why divers use helium to avoid getting sick.

Exam Tip: State the mathematical relationship \( p = K_H \cdot x \) alongside the verbal definition, and mention scuba diving and carbonated drinks as standard applications.

Question. Why do gases nearly always tend to be less soluble in liquids as the temperature is raised?
Answer: This occurs because the dissolution of a gas in a liquid is an exothermic reaction. According to Le Chatelier's principle, solubility must decline when temperature rises.
In simple words: When gas dissolves in liquid, it releases heat. If you add more heat by warming the liquid, the system fights back by pushing the gas out, reducing solubility.

Exam Tip: Use Le Chatelier's principle and explicitly state that the gas dissolution process is exothermic (\( \Delta H < 0 \)) to get full marks.

Question. 18 g of glucose, C6H₁₂O₆ (Molar mass = 180 g mol^–1) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil? (Kb for water = 0.512 K kg mol^–1, boiling point of pure water = 373.15 K)
Answer: We know that the elevation of boiling point is: \[ \Delta T_b = \frac{W_B \times 1000 \times K_b}{M_B \times W_A} \] Given: Mass of solute (glucose), \( W_B = 18\text{ g} \) Molar mass of glucose, \( M_B = 180\text{ g mol}^{-1} \) Mass of solvent (water), \( W_A = 1\text{ kg} = 1000\text{ g} \) \( K_b \) of water = \( 0.512\text{ K kg mol}^{-1} \) Substituting these values: \( \Delta T_b = \frac{18 \times 1000 \times 0.512}{180 \times 1000} \)
\( \implies \Delta T_b = 0.052\text{ K} \) Boiling point of the solution: \( T_b = T_b^0 + \Delta T_b = 373.15\text{ K} + 0.052\text{ K} = 373.202\text{ K} \) Therefore, the solution will boil at \( 373.202\text{ K} \).
In simple words: Adding glucose raises water's boiling point slightly. We calculate this increase to be 0.052 Kelvin, making the final boiling temperature 373.202 Kelvin.

Exam Tip: Be careful to add the elevation (\( \Delta T_b \)) to the boiling point of pure water (\( 373.15\text{ K} \)) to find the final boiling point of the solution.

Question. Derive expression for Raoult’s law when the solute is non-volatile.
Answer: For a solution containing a non-volatile solute, the vapour pressure is solely due to the solvent molecules. According to Raoult's law, the partial vapour pressure of the solvent (\( p_1 \)) in a solution is directly proportional to its mole fraction (\( X_1 \)). Mathematically: \[ p_1 = p_1^0 \cdot X_1 \] Where \( p_1^0 \) represents the vapour pressure of the pure solvent. Since it is a binary system, the mole fractions of solvent (\( X_1 \)) and solute (\( X_2 \)) satisfy: \( X_1 + X_2 = 1 \implies X_1 = 1 - X_2 \) Replacing \( X_1 \): \( p_1 = p_1^0 (1 - X_2) \)
\( \implies p_1 = p_1^0 - p_1^0 X_2 \)
\( \implies p_1^0 X_2 = p_1^0 - p_1 \)
\( \implies X_2 = \frac{p_1^0 - p_1}{p_1^0} \) Here, \( \frac{p_1^0 - p_1}{p_1^0} \) is the relative lowering of vapour pressure, which is equal to the mole fraction of the non-volatile solute.
In simple words: By swapping the solvent's mole fraction for (1 - solute's fraction), we show that the percentage drop in vapor pressure is exactly equal to the ratio of solute particles added.

Exam Tip: Finish your derivation by explicitly naming the term \( \frac{p_1^0 - p_1}{p_1^0} \) as "relative lowering of vapour pressure" to earn full marks.

Question. Calculate the mass of compound (molar mass = 256 g mol^–1) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K (Kf = 5.12 K kg mol^–1).
Answer: We use the formula for depression in freezing point: \[ \Delta T_f = \frac{K_f \times W_2 \times 1000}{M_2 \times W_1} \] Rearranging to solve for the mass of solute (\( W_2 \)): \[ W_2 = \frac{M_2 \times W_1 \times \Delta T_f}{1000 \times K_f} \] Given: \( \Delta T_f = 0.48\text{ K} \) Molar mass of solute, \( M_2 = 256\text{ g mol}^{-1} \) Mass of solvent (benzene), \( W_1 = 75\text{ g} \) \( K_f \) for benzene = \( 5.12\text{ K kg mol}^{-1} \) Substituting these values: \( W_2 = \frac{256 \times 75 \times 0.48}{1000 \times 5.12} \)
\( \implies W_2 = 1.8\text{ g} \) Thus, the required mass of the compound is \( 1.8\text{ g} \).
In simple words: Rearranging the freezing point formula let us calculate that 1.8 grams of the compound are needed to lower benzene's freezing point by 0.48 Kelvin.

Exam Tip: Write down the formula in its standard form first before rearranging it, and clearly list the given variables with their correct units.

Question. Define an ideal solution and write one of its characteristics.
Answer: An ideal solution is one that obeys Raoult's law across the entire range of concentration and temperature. Characteristics: 1. There is no enthalpy change upon mixing: \( \Delta H_{\text{mixing}} = 0 \). 2. There is no change in volume during mixing: \( \Delta V_{\text{mixing}} = 0 \).
In simple words: An ideal solution obeys Raoult's law perfectly. Mixing its parts doesn't release or absorb any heat, and the total volume is exactly equal to the sum of the parts.

Exam Tip: Remember to state both thermodynamic conditions (\( \Delta H_{\text{mix}} = 0 \) and \( \Delta V_{\text{mix}} = 0 \)) when describing the characteristics of an ideal solution.

Question. State Henry’s law. What is the effect of temperature on the solubility of a gas in a liquid?
Answer: Henry’s law states that the mass of a gas dissolved in a given volume of liquid is directly proportional to the pressure of the gas in dynamic equilibrium with the liquid at a constant temperature. As temperature increases, the solubility of a gas in a liquid decreases due to the exothermic nature of the dissolution process.
In simple words: Henry's law says that pushing gas harder dissolves more of it in liquid. Warming up the liquid makes the gas escape, lowering its solubility.

Exam Tip: Explain the temperature effect clearly by noting that gas dissolution is an exothermic process, which is why higher temperatures favor the gas escaping.

Question. State Raoult’s law for the solution containing volatile components. What is the similarity between Raoult’s law and Henry’s law?
Answer: Raoult's law for volatile systems states that the partial vapour pressure of any component at a particular temperature is equivalent to its mole fraction in the mixture multiplied by its pure state vapour pressure. The main similarity between Raoult’s law and Henry’s law is that both equations assert that the partial pressure of a volatile component is directly proportional to its mole fraction in the solution.
In simple words: Both laws state that the pressure of a vapor is directly proportional to how much of that substance is in the liquid mixture.

Exam Tip: Underline that in both laws, partial pressure is proportional to mole fraction, with the only difference being the proportionality constant (\( p^\circ \) for Raoult and \( K_H \) for Henry).

Question. How is the vapour pressure of a solvent affected when a non-volatile solute is dissolved in it?
Answer: The vapour pressure of a solvent drops when a non-volatile solute is introduced. This happens because some of the surface area is occupied by solute particles, reducing the number of escaping solvent molecules.
In simple words: Adding non-volatile solids to a liquid blocks part of the surface, making it harder for liquid molecules to evaporate and lowering the vapor pressure.

Exam Tip: Focus your explanation on the reduction of the liquid's surface area available for solvent molecules to escape into the vapor phase.

Question. Differentiate between molarity and molality of a solution. How can we change molality value of a solution into molarity value?
Answer: Molarity (\( M \)) is the moles of solute contained in one liter of solution, whereas molality (\( m \)) is the moles of solute per kilogram of solvent. To convert molality into molarity, the density (\( d \)) of the solution must be known. If \( M \) is the molarity and \( M_2 \) is the molar mass of the solute: Mass of \( 1000\text{ mL} \) of solution = \( 1000d\text{ g} \) Mass of solute present = \( M \cdot M_2\text{ g} \) Mass of solvent = \( (1000d - M \cdot M_2)\text{ g} \) By definition, molality is: \[ m = \frac{M \times 1000}{1000d - M \cdot M_2} \] This can be rearranged to find molarity (\( M \)): \[ M = \frac{m \cdot d}{1 + \frac{m \cdot M_2}{1000}} \]
In simple words: Molarity uses the liquid volume, while molality uses the solvent's weight. To convert between them, you must use the solution's density to connect volume to weight.

Exam Tip: Remember to specify that density is the essential link required to convert between volume-based molarity and mass-based molality.

Short Answer Type Questions-II (SA-II)

Question. What is meant by positive deviations from Raoult’s law? Give an example. What is the sign of ΔmixH for positive deviation?
Answer: Positive deviations from Raoult’s law occur when the observed vapour pressure of each individual component and the entire solution is higher than the values calculated theoretically. An example is a mixture of water and ethanol. For these solutions, the enthalpy of mixing is positive: \( \Delta_{\text{mix}}H > 0 \).
In simple words: Positive deviation means the liquid mixture boils easier than expected because the different molecules don't stick to each other very well. This mixing absorbs heat, making enthalpy positive.

Exam Tip: Specify that \( \Delta_{\text{mix}}H \) is positive (greater than zero) because breaking stronger solute-solute and solvent-solvent interactions requires absorbing energy.

Question. Define azeotropes. What type of azeotrope is formed by positive deviation from Raoult’s law? Given an example.
Answer: Azeotropes: These are liquid mixtures that boil at a constant temperature and distill without any alteration in their composition. Solutions showing large positive deviations from Raoult’s law form minimum boiling azeotropes. A classic example is a mixture containing 95% ethanol and 5% water by volume.
In simple words: Azeotropes are mixtures that act like a pure liquid when boiling. Large positive deviations lead to minimum boiling azeotropes, which boil at a lower temperature than either pure component.

Exam Tip: Note that large positive deviations lead to minimum boiling azeotropes because the higher-than-expected vapour pressure allows the mixture to boil at a lower temperature.

Question. (i) On mixing liquid X and liquid Y, volume of the resulting solution decreases. What type of deviation from Raoult’s law is shown by the resulting solution? What change in temperature would you observe after mixing liquids X and Y? (ii) What happens when we place the blood cell in water (hypotonic solution)? Give reason.
Answer: (i) Since the total volume shrinks upon mixing, this indicates negative deviation from Raoult's law. Because the process is exothermic (\( \Delta H_{\text{mix}} < 0 \")), a rise in temperature of the mixture will be observed.
(ii) When placed in a hypotonic solution like water, the blood cell will swell and may eventually burst. This occurs because water enters the cell through osmosis.
In simple words: (i) A smaller volume means the molecules pull closer together, releasing heat and warming up the mixture. (ii) Water flows into the salty blood cell, making it inflate and pop.

Exam Tip: For part (i), explicitly state that a volume decrease (\( \Delta V_{\text{mix}} < 0 \)) corresponds to a negative deviation, which is always exothermic (\( \Delta H_{\text{mix}} < 0 \)).

Question. Define osmotic pressure of a solution. How is the osmotic pressure related to the concentration of a solute in a solution?
Answer: Osmotic pressure: This is the external pressure applied to the solution to halt the flow of solvent across a semipermeable membrane. Relationship to concentration: The osmotic pressure (\( \pi \)) is directly proportional to the molar concentration (\( C \)) of the solute at a given temperature: \[ \pi = C \cdot R \cdot T \] Where \( R \) is the gas constant and \( T \) is the absolute temperature. Since concentration \( C = \frac{n}{V} \), we can also write: \[ \pi = \frac{n}{V}RT \]
In simple words: Osmotic pressure is the squeeze needed to stop pure water from invading a solution. It goes up directly with the concentration of solute dissolved in the liquid.

Exam Tip: Show the step-by-step formula derivation connecting osmotic pressure (\( \pi \)) to concentration (\( C \)) and molar mass using \( C = \frac{n}{V} \).

Question. Define the following terms : (i) Mole fraction (x) (ii) Molality of a solution (m)
Answer: (i) Mole fraction (\( x \)): It represents the ratio of the moles of a specific component to the sum of the moles of all components in the mixture. For a binary mixture: \[ x_1 = \frac{n_1}{n_1 + n_2} \] (ii) Molality (\( m \)): This refers to the total moles of solute dissolved in 1000 grams (or one kilogram) of the solvent. It is calculated using: \[ m = \frac{w \times 1000}{M \times W} \] Where \( w \) is the mass of solute, \( M \) is its molar mass, and \( W \) is the mass of solvent.
In simple words: Mole fraction is the fraction of molecules that belong to one component. Molality is simply how many moles of solute you mixed into a kilogram of solvent.

Exam Tip: Be sure to write the full mathematical formulas when defining mole fraction and molality to secure full marks.

Question. (i) Gas (A) is more soluble in water than Gas (B) at the same temperature. Which one of the two gases will have the higher value of KH (Henry’s constant) and why? (ii) In non-ideal solution, what type of deviation shows the formation of maximum boiling azeotropes?
Answer: (i) Gas (B) will possess a higher Henry's law constant (\( K_H \)) value than Gas (A). This is because the solubility of a gas is inversely proportional to its Henry's constant, meaning lower solubility corresponds to a larger \( K_H \).
(ii) Non-ideal solutions exhibiting large negative deviations from Raoult's law result in maximum boiling azeotropes.
In simple words: (i) Since Gas B is less soluble, its Henry's constant is higher because solubility and the constant are opposites. (ii) Large negative deviations create maximum boiling azeotropes, which boil at a higher temperature.

Exam Tip: State the relation \( p = K_H \cdot x \) to prove that higher \( K_H \) values lead to lower mole fractions (solubility) of the gas in solution at a given pressure.

Question. What is osmotic pressure? Why it is a colligative property?
Answer: Osmotic pressure is the minimum external pressure that must be applied to the solution to prevent osmosis. It is classified as a colligative property because its value depends solely on the concentration of solute particles, not on their identity.
In simple words: Osmotic pressure stops osmosis from happening. It is a colligative property because only the number of dissolved particles matters, not what kind of particles they are.

Exam Tip: Always define colligative properties as those depending strictly on the number (concentration) of solute particles rather than their chemical nature.

Question. Define osmotic pressure. How is osmotic pressure related to the concentration of a solute in a solution?
Answer: Osmotic pressure is the external pressure required to prevent the inflow of solvent molecules across a semipermeable membrane. It is directly proportional to the molar concentration of the solute in the solution, expressed as \( \pi = CRT \).
In simple words: Osmotic pressure is the force needed to block solvent molecules. It rises directly as the solution gets more concentrated.

Exam Tip: Be sure to write the formula \( \pi = CRT \) and define each term (\( C \) = molarity, \( R \) = gas constant, \( T \) = temperature) to receive full marks.

Question. Define the following terms: (i) Colligative properties (ii) Molality (m)
Answer: (i) Colligative properties: These are physical properties of solutions that depend solely on the total number of solute particles present, regardless of their nature or chemical identity.
(ii) Molality (\( m \)): This is defined as the number of moles of solute dissolved in one kilogram of the solvent.
In simple words: (i) Colligative properties care only about "how many" particles are dissolved, not "what" they are. (ii) Molality is moles of solute dissolved in 1 kg of solvent.

Exam Tip: Ensure that you clearly state the distinction between solvent mass (used in molality) and solution volume (used in molarity).

Question. Define the following terms: (i) Abnormal molar mass (ii) van’t Hoff factor (i)
Answer: (i) Abnormal molar mass: When the molecular mass of a solute determined experimentally via colligative properties deviates from its actual theoretical value, it is referred to as abnormal molar mass.
(ii) van't Hoff factor (\( i \)): This represents the degree of association or dissociation of solute particles, defined as the ratio of the experimental value of a colligative property to its theoretical value: \[ i = \frac{\text{Observed colligative property}}{\text{Theoretical colligative property}} \]
In simple words: Abnormal molar mass is when a dissolved substance clumps or splits, making it weigh differently than expected. The van't Hoff factor is the ratio measuring this change.

Exam Tip: State that abnormal molar masses occur when solutes undergo association (clumping) or dissociation (splitting) in a solvent.

Question. Define the following terms: (i) Ideal solution (ii) Molarity (M)
Answer: (i) Ideal solution: A solution that obeys Raoult's law under all concentrations and temperatures is called an ideal solution.
(ii) Molarity (\( M \)): This represents the number of solute moles dissolved per liter of solution, computed as: \[ M = \frac{W_B \times 1000}{M_B \times V_{\text{mL}}} \]
In simple words: An ideal solution is a perfect mixture obeying Raoult's law. Molarity measures how many moles of solute are dissolved in one liter of total solution.

Exam Tip: Always specify "per liter of solution" for molarity and write the mathematical equation with all units clearly marked.

Question. Explain why on addition of 1 mol of glucose to 1 litre of water, the boiling point of water increases.
Answer: When a non-volatile solute like glucose is dissolved in water, the vapour pressure of the liquid decreases because solute particles occupy space on the surface. Consequently, a higher temperature is required to raise its vapour pressure to match the atmospheric pressure, causing an elevation in the boiling point.
In simple words: Glucose particles block the surface of the water, making it harder for water to evaporate. To get it to boil, you must heat it up more, which raises the boiling point.

Exam Tip: Connect the reduction in vapour pressure directly to the increase in boiling point temperature, as boiling happens only when vapour pressure equals atmospheric pressure.

Question 38. 100 mg of a protein is dissolved in just enough water to make 10.0 mL of solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25°C, what is the molar mass of the protein? (R = 0.0821 L atm mol^–1 K^–1 and 760 mm Hg = 1 atm.)
Answer: We use the osmotic pressure formula: \[ \pi = \frac{wRT}{MV} \implies M = \frac{wRT}{\pi V} \] Given: Mass of protein, \( w = 100\text{ mg} = 0.1\text{ g} \) Volume of solution, \( V = 10.0\text{ mL} = 0.01\text{ L} \) Osmotic pressure, \( \pi = \frac{13.3}{760}\text{ atm} \) Temperature, \( T = 25^\circ\text{C} = 25 + 273 = 298\text{ K} \) Gas constant, \( R = 0.0821\text{ L atm mol}^{-1}\text{ K}^{-1} \) Substituting these values: \[ M = \frac{0.1 \times 0.0821 \times 298}{\left(\frac{13.3}{760}\right) \times 0.01} \] \[ M = \frac{0.1 \times 0.0821 \times 298 \times 760}{13.3 \times 0.01} \]
\( \implies M = \frac{1859.4008}{0.133} \approx 13980.46\text{ g mol}^{-1} \) Thus, the molar mass of the protein is \( 13980.46\text{ g mol}^{-1} \).
In simple words: Convert the protein mass to grams and pressure to atmospheres, then plug these into the osmotic pressure formula to find that the molar mass of the protein is 13980.46 grams per mole.

Exam Tip: Convert pressure from mm Hg to atm and volume from mL to L before substituting them into the formula to ensure unit consistency with the gas constant \( R \).

Question. Calculate the freezing point depression expected for 0.0711 m aqueous solution of Na2SO4. If this solution actually freezes at – 0.320°C, what would be the value of Van’t Hoff factor? (Kf for water is 1.86°C mol^–1)
Answer: The observed depression in freezing point is: \( \Delta T_f = T_f^\circ - T_f = 0^\circ\text{C} - (-0.320^\circ\text{C}) = 0.320^\circ\text{C} = 0.320\text{ K} \) We know that: \[ \Delta T_f = i \cdot K_f \cdot m \] Rearranging for the van't Hoff factor (\( i \)): \[ i = \frac{\Delta T_f}{K_f \cdot m} \] Given: \( m = 0.0711\text{ m} \) \( K_f = 1.86\text{ K kg mol}^{-1} \) Substituting these values: \( i = \frac{0.320}{1.86 \times 0.0711} \)
\( \implies i = \frac{0.320}{0.132246} \approx 2.42 \) Thus, the value of the van't Hoff factor is \( 2.42 \).
In simple words: Since the freezing point drops by 0.320 degrees, we divide this by the freezing point constant and molality to find a van't Hoff factor of 2.42.

Exam Tip: Remember to express the depression in freezing point (\( \Delta T_f \)) as a positive temperature difference, representing how much the freezing point has lowered.

Question. A solution prepared by dissolving 1.25 g of oil of winter green (methyl salicylate) in 99.0 g of benzene has a boiling point of 80.31°C. Determine the molar mass of this compound. (B.P. of pure benzene = 80.10°C and Kb for benzene = 2.53°C kg mol^–1).
Answer: We use the formula for the elevation in boiling point: \[ \Delta T_b = \frac{K_b \times W_2 \times 1000}{M_2 \times W_1} \] Rearranging for the molar mass of the solute (\( M_2 \)): \[ M_2 = \frac{1000 \times K_b \times W_2}{W_1 \times \Delta T_b} \] Given: Mass of solute, \( W_2 = 1.25\text{ g} \) Mass of solvent, \( W_1 = 99.0\text{ g} \) \( \Delta T_b = 80.31^\circ\text{C} - 80.10^\circ\text{C} = 0.21^\circ\text{C} = 0.21\text{ K} \) \( K_b \) of benzene = \( 2.53\text{ K kg mol}^{-1} \) Substituting these values: \( M_2 = \frac{1000 \times 2.53 \times 1.25}{99.0 \times 0.21} \)
\( \implies M_2 = \frac{3162.5}{20.79} \approx 152.12\text{ g mol}^{-1} \) Thus, the molar mass of the compound is \( 152.12\text{ g mol}^{-1} \).
In simple words: The temperature rose by 0.21 degrees when the oil of wintergreen was dissolved. Rearranging the formula tells us its molar mass is 152.12 grams per mole.

Exam Tip: Always double check the boiling point elevation (\( \Delta T_b \)) by subtracting pure solvent's boiling point from that of the solution.

Question. A solution of glycerol (C3H8O3 ; molar mass = 92 g mol–1) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42 °C. What mass of glycerol was dissolved to make this solution? Kb for water = 0.512 K kg mol^–1.
Answer: We use the boiling point elevation formula: \[ \Delta T_b = \frac{K_b \times W_2 \times 1000}{M_2 \times W_1} \] Rearranging for the mass of the solute (\( W_2 \)): \[ W_2 = \frac{W_1 \times M_2 \times \Delta T_b}{1000 \times K_b} \] Given: Molar mass of glycerol (\( M_2 \)) = \( 92\text{ g mol}^{-1} \) Mass of solvent (water), \( W_1 = 500\text{ g} \) Elevation in boiling point, \( \Delta T_b = 100.42^\circ\text{C} - 100^\circ\text{C} = 0.42^\circ\text{C} = 0.42\text{ K} \) \( K_b \) for water = \( 0.512\text{ K kg mol}^{-1} \) Substituting these values: \( W_2 = \frac{500 \times 92 \times 0.42}{1000 \times 0.512} \)
\( \implies W_2 = \frac{19320}{512} \approx 37.73\text{ g} \) Thus, the mass of glycerol dissolved is \( 37.73\text{ g} \).
In simple words: Rearranging the boiling point formula reveals that 37.73 grams of glycerol are needed to raise the boiling point of 500 grams of water by 0.42 Kelvin.

Exam Tip: Ensure that the boiling point elevation is computed as \( T_{\text{solution}} - 100^\circ\text{C} \) because water's normal boiling point is precisely \( 100^\circ\text{C} \).