Maharashtra Board Class 12 Chemistry Chapter 2 Solutions

Choose the Most Correct Answer.

Question i. The vapour pressure of a solution containing 2 moles of a solute in 2 moles of water (vapour pressure of pure water = 24 mm Hg) is
(a) 24 mm Hg
(b) 32 mm Hg
(c) 48 mm Hg
(d) 12 mm Hg
Answer: (d) 12 mm Hg
In simple words: Since half of the mixture is solute and half is water, the mole fraction of water is 0.5. According to Raoult's law, the new vapour pressure is half of the pure water's pressure, which is 12 mm Hg.

🎯 Exam Tip: Always calculate the mole fraction of the solvent (not the solute) when using Raoult's law to find the vapour pressure of a solution.

Question ii. The colligative property of a solution is
(a) vapour pressure
(b) boiling point
(c) osmotic pressure
(d) freezing point
Answer: (c) osmotic pressure
In simple words: Colligative properties depend only on the number of solute particles. Osmotic pressure is a colligative property, whereas simple boiling point or freezing point are just physical properties of pure substances.

🎯 Exam Tip: Examiners often try to trick you by listing "boiling point" instead of "elevation of boiling point". Only the elevation, depression, or relative lowering are colligative properties, except for osmotic pressure which is a colligative property on its own.

Question iii. In calculating osmotic pressure the concentration of solute is expressed in
(a) molarity
(b) molality
(c) mole fraction
(d) mass per cent
Answer: (a) molarity
In simple words: Osmotic pressure is measured at a fixed temperature, so we express the concentration in molarity (moles per litre) to easily relate it to the volume of the solution.

🎯 Exam Tip: Remember the formula \( \pi = CRT \), where \( C \) stands for molarity. This is a very common formula for numerical problems.

Question iv. Ebullioscopic constant is the boiling point elevation when the concentration of solution is
(a) 1 m
(b) 1 M
(c) 1 mass%
(d) 1 mole fraction of solute
Answer: (a) 1 m
In simple words: The ebullioscopic constant is the rise in boiling point when we dissolve exactly 1 mole of solute in 1 kilogram of solvent, which is a 1 molal (1 m) solution.

🎯 Exam Tip: Remember that ebullioscopic constant (\( K_b \)) is defined per unit molality (lowercase 'm'), not molarity (uppercase 'M').

Question v. Cryoscopic constant depends on
(a) nature of solvent
(b) nature of solute
(c) nature of solution
(d) number of solvent molecules
Answer: (a) nature of solvent
In simple words: The cryoscopic constant is a fixed property of the liquid solvent itself and does not change based on what solute you dissolve in it.

🎯 Exam Tip: Both cryoscopic (\( K_f \)) and ebullioscopic (\( K_b \)) constants are characteristic properties of the solvent only, determined by its molecular weight and temperatures.

Question vi. Identify the correct statement
(a) vapour pressure of solution is higher than that of pure solvent.
(b) boiling point of solvent is lower than that of solution
(c) osmotic pressure of solution is lower than that of solvent
(d) osmosis is a colligative property.
Answer: (b) boiling point of solvent is lower than that of solution
In simple words: When you add a solute to a liquid, its boiling point goes up. This means the pure solvent always boils at a lower temperature than the solution.

🎯 Exam Tip: Be careful with option (d); "osmotic pressure" is the colligative property, whereas "osmosis" is just the physical process itself.

Question vii. A living cell contains a solution which is isotonic with 0.3 M sugar solution. What osmotic pressure develops when the cell is placed in 0.1 M KCl solution at body temperature ?
(a) 5.08 atm
(b) 2.54 atm
(c) 4.92 atm
(d) 2.46 atm
Answer: (b) 2.54 atm
In simple words: The cell has a concentration of 0.3 M, while the KCl solution outside splits into ions to act like a 0.2 M solution. The difference of 0.1 M creates an osmotic pressure of 2.54 atm at body temperature.

🎯 Exam Tip: Always remember to multiply the concentration of strong electrolytes like \( \text{KCl} \) by their van 't Hoff factor (\( i = 2 \)) before calculating osmotic pressure differences.

Question viii. The osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose isotonic with blood has the percentage (by volume)
(a) 5.41%
(b) 3.54%
(c) 4.53%
(d) 53.4%
Answer: (a) 5.41%
In simple words: Isotonic solutions have the same osmotic pressure. A 5.41% glucose solution has the exact same osmotic pressure as blood, making them compatible.

🎯 Exam Tip: Remember that isotonic solutions have equal molar concentrations. Use the osmotic pressure formula \( \pi = CRT \) to find the concentration when comparing solutions.

Question ix. Vapour pressure of a solution is
(a) directly proportional to the mole fraction of the solute
(b) inversely proportional to the mole fraction of the solute
(c) inversely proportional to the mole fraction of the solvent
(d) directly proportional to the mole fraction of the solvent
Answer: (d) directly proportional to the mole fraction of the solvent
In simple words: According to Raoult's law, more solvent molecules on the surface means more of them can escape into the air, which directly increases the vapour pressure.

🎯 Exam Tip: Always associate Raoult's Law directly with the mole fraction of the solvent, as the vapour pressure of the solution depends on the active solvent molecules at the surface.

Question x. Pressure cooker reduces cooking time for food because
(a) boiling point of water involved in cooking is increased
(b) heat is more evenly distributed in the cooking space
(c) the higher pressure inside the cooker crushes the food material
(d) cooking involves chemical changes helped by a rise in temperature
Answer: (a) boiling point of water involved in cooking is increased
In simple words: Inside a pressure cooker, the high pressure prevents water from boiling at its usual temperature, letting it get much hotter so the food cooks much faster.

🎯 Exam Tip: Remember that boiling point increases with an increase in external pressure. This is a classic application of colligative properties and phase equilibrium.

Question xi. Henry’s law constant for a gas \( \text{CH}_3\text{Br} \) is \( 0.159 \text{ mol dm}^{-3}\text{ atm}^{-1} \) at \( 25^\circ\text{C} \). What is the solubility of \( \text{CH}_3\text{Br} \) in water at \( 25^\circ\text{C} \) and a partial pressure of \( 0.164 \text{ atm} \)?
(a) \( 0.0159 \text{ mol L}^{-1} \)
(b) \( 0.164 \text{ mol L}^{-1} \)
(c) 0.026 M
(d) 0.042 M
Answer: (c) 0.026 M
In simple words: To find the solubility, we simply multiply Henry's constant by the pressure of the gas, which gives us \( 0.159 \times 0.164 = 0.026 \text{ M} \).

🎯 Exam Tip: Always check that the units of Henry's constant and pressure match before multiplying them directly using the formula \( S = K_H \times P \).

Question xii. Which of the following statement is NOT correct for 0.1 M urea solution and 0.05 M sucrose solution ?
(a) osmotic pressure exhibited by urea solution is higher than that exhibited by sucrose solution
(b) urea solution is hypertonic to sucrose solution
(c) they are isotonic solutions
(d) sucrose solution is hypotonic to urea solution
Answer: (c) they are isotonic solutions
In simple words: Since the concentration of the urea solution (0.1 M) is different from the sucrose solution (0.05 M), they cannot have the same osmotic pressure, meaning they are not isotonic.

🎯 Exam Tip: Remember that isotonic solutions must have the same molar concentration at a given temperature. Since 0.1 M and 0.05 M are different, they cannot be isotonic.

Answer the Following in One or Two Sentences

Question i. What is osmotic pressure ?
Answer: (1) Definition : The osmotic pressure is defined as the excess mechanical pressure required to be applied to a solution separated by a semipermeable membrane from pure solvent or a dilute solution to prevent the osmosis or free passage of the solvent molecules at a given temperature. The osmotic pressure is a colligative property. This property depends solely on the number of solute particles in the solution.
In simple words: Osmotic pressure is the extra pressure you need to apply to a solution to stop the natural flow of solvent molecules through a semipermeable membrane.

🎯 Exam Tip: Always define osmotic pressure clearly and mention that it is a colligative property to secure full marks.

Osmosis and Osmotic Pressure

Diagram Labels:

• Hydrostatic pressure (hg)
• Sugar solution
• Semipermeable membrane
• Water

Explanation: Consider an inverted thistle funnel on the mouth of which a semipermeable membrane is firmly fastened. It is filled with the experimental solution and immersed in a solvent like water. As a result, solvent molecules pass through the membrane into the solution in the funnel causing rising of level in the arm of thistle funnel. This increases the hydrostatic pressure. At a certain stage this rising level stops indicating an equilibrium between the rates of flow of solvent molecules from solvent to solution and from solution to solvent. The hydrostatic pressure at this stage represents osmotic pressure of the solution in the thistle funnel.

Question ii. A solution concentration is expressed in molarity and not in molality while considering osmotic pressure. Why?
Answer:
1. While calculating osmotic pressure by equation, \( \pi = CRT \), the concentration is expressed in molarity but not in molality.
2. This is because the measurements of osmotic pressure are made at a certain constant temperature.
3. Molarity depends upon temperature but molality is independent of temperature.
4. Hence in osmotic pressure measurements, concentration is expressed in molarity, which relates directly to the volume of the solution at that specific temperature.
In simple words: Osmotic pressure experiments are conducted at a constant temperature. Since molarity depends on volume (which changes with temperature) and the temperature is kept constant during the measurement, molarity is the preferred and more practical unit to use.

🎯 Exam Tip: Always write down the formula \( \pi = CRT \) and explicitly mention that molarity is temperature-dependent while molality is temperature-independent to secure full marks.

Question iii. Write the equation relating boiling point elevation to the concentration of solution.
Answer: The elevation in the boiling point of a solution is directly proportional to the molal concentration (expressed in \( \text{mol kg}^{-1} \)) of the solution. This relationship highlights how adding a non-volatile solute restricts the solvent molecules from escaping into the vapor phase easily.
Hence, if \( \Delta T_b \) is the elevation in the boiling point of a solution of molal concentration \( m \) then,
\( \Delta T_b \propto m \)
\( \therefore \Delta T_b = K_b m \)
where \( K_b \) is a proportionality constant.
If \( m = 1 \) molal,
\( \Delta T_b = K_b \)
\( K_b \) is called the ebullioscopic constant or molal elevation constant. \( K_b \) is characteristic of the solvent.
In simple words: When you dissolve something in a liquid, its boiling point goes up. This increase in boiling temperature is directly related to how much solute you added to the liquid.

🎯 Exam Tip: Always define \( K_b \) (ebullioscopic constant) and state its units (\( \text{K kg mol}^{-1} \)) clearly to secure full marks in descriptive questions.

Question iv. A \( 0.1\text{ m} \) solution of \( \text{K}_2\text{SO}_4 \) in water has freezing point of \( -0.43\text{ }^\circ\text{C} \). What is the value of van’t Hoff factor if \( K_f \) for water is \( 1.86\text{ K kg mol}^{-1} \)?
Answer: Given data:
\( m = 0.1\text{ m} \)
\( \Delta T_f = 0 - (-0.43) = 0.43\text{ }^\circ\text{C} \)
\( K_f = 1.86\text{ K kg mol}^{-1} \)
\( i = ? \)

Formula:
\( \Delta T_f = i \times K_f \times m \)
\( \therefore i = \frac{\Delta T_f}{K_f \times m} \)
\( \therefore i = \frac{0.43}{1.86 \times 0.1} \)
\( \therefore i = 2.312 \)
The calculated value shows that the solute undergoes partial dissociation in the solution.
Thus, the van’t Hoff factor \( i = 2.312 \).
In simple words: The van't Hoff factor tells us how many particles a substance splits into when dissolved. Here, the value of 2.312 shows that the salt partially breaks down into ions in water.

🎯 Exam Tip: Double-check your calculations for \( \Delta T_f \) by ensuring you subtract the solution's freezing point from the pure solvent's freezing point (\( 0\text{ }^\circ\text{C} \)).

Question v. What is van’t Hoff factor?
Answer: The van’t Hoff factor (\( i \)) is defined as the ratio of the observed colligative property of the solution to the theoretically calculated colligative property of the solution without considering molecular change. It helps us understand whether a solute associates or dissociates when dissolved in a solvent.
The van’t Hoff factor can be represented as:
\[ i = \frac{\text{Observed value of colligative property}}{\text{Theoretical value of the colligative property}} \]
In simple words: The van't Hoff factor is a ratio that compares the actual measured properties of a solution to what we theoretically expect if the solute particles did not split or clump together.

🎯 Exam Tip: Write down the mathematical formula alongside the verbal definition to make your answer highly structured and easy to grade.

Van't Hoff Factor and Colligative Properties

This colligative property may be the lowering of vapour pressure of a solution, the osmotic pressure, the elevation in the boiling point or the depression in the freezing point of the solution. Hence,

\( i = \frac{\text{Observed lowering of vapour pressure}}{\text{Theoretical lowering of vapour pressure}} = \frac{\Delta P_{\text{(ob)}}}{\Delta P_{\text{(th)}}} \)

\( i = \frac{\text{Observed elevation in boiling point}}{\text{Theoretical elevation in boiling point}} = \frac{\Delta T_{\text{b(ob)}}}{\Delta T_{\text{b(th)}}} \)

\( i = \frac{\text{Observed depression in freezing point}}{\text{Theoretical depression in freezing point}} = \frac{\Delta T_{\text{f(ob)}}}{\Delta T_{\text{f(th)}}} \)

\( i = \frac{\text{Observed osmotic pressure}}{\text{Theoretical osmotic pressure}} = \frac{\pi_{\text{(ob)}}}{\pi_{\text{(th)}}} \)

• When the solute neither undergoes dissociation or association in the solution, then, \( i = 1 \)
• When the solute undergoes dissociation in the solution, then, \( i > 1 \)
• When the solute undergoes association in the solution, then \( i < 1 \)

From the value of the van’t Hoff factor, the degree of dissociation of electrolytes, degree of association of nonelectrolytes can be obtained.

van’t Hoff factor gives the important information about the solute molecules in the solution and chemical bonding in them.

Question vi. How is van’t Hoff factor related to degree of ionization?
Answer: Consider \( 1 \text{ dm}^3 \) of a solution containing \( m \) moles of an electrolyte \( A_xB_y \). The electrolyte on dissociation gives \( x \) number of \( A^{y+} \) ions and \( y \) number of \( B^{x-} \) ions. Let \( \alpha \) be the degree of dissociation. At equilibrium, \( A_xB_y \rightleftharpoons xA^{y+} + yB^{x-} \). This relationship helps us determine the extent to which a substance breaks down into ions in a solution.
In simple words: The van't Hoff factor tells us how many particles are actually in the solution compared to what we put in. By knowing this, we can calculate the percentage of molecules that split into ions.

🎯 Exam Tip: Clearly define all terms like \( \alpha \) (degree of dissociation) and write the balanced equilibrium equation to secure full marks in derivation questions.

Relation Between Van 't Hoff Factor and Degree of Dissociation

For 1 mole of electrolyte: \( 1 - \alpha \), \( x\alpha \), \( y\alpha \)

and For ‘m’ moles of an electrolyte: \( m(1 - \alpha) \), \( mx\alpha \), \( my\alpha \) are the number of particles.

Total number of moles at equilibrium, will be,
Total moles = \( m(1 - \alpha) + mx\alpha + my\alpha \)
\( = m[(1 - \alpha) + x\alpha + y\alpha] \)
\( = m[1 + x\alpha + y\alpha - \alpha] \)
\( = m[1 + \alpha(x + y - 1)] \)

The van’t Hoff factor \( i \) will be,
\( i = \frac{\text{Observed colligative property}}{\text{Theoretical colligative property}} \)
\( = \frac{m[1 + \alpha(x + y - 1)]}{m} \)
\( i = 1 + \alpha(x + y - 1) \)

If total number of ions from one mole of electrolyte is denoted by \( n \), then \( (x + y) = n \)
\( \therefore i = 1 + \alpha(n - 1) \)

\( \implies \alpha(n - 1) = i - 1 \)

\( \implies \alpha = \frac{i - 1}{n - 1} \) ......(1)

This is a relation between van’t Hoff factor \( i \) and degree of dissociation of an electrolyte.

Question vii. Which of the following solutions will have higher freezing point depression and why?
a. 0.1 m NaCl
b. 0.05 m Al₂(SO₄)₃
Answer:
(1) Freezing point depression is a colligative property, hence depends on the number of particles in the solution.
(2) More the number of particles in the solution, higher is the depression in freezing point.
(3) The number of particles (ions) from electrolytes are:
• For 0.1 m NaCl: NaCl dissociates into 2 ions (\(\text{Na}^+\) and \(\text{Cl}^-\)). Total concentration of particles = \(2 \times 0.1\text{ m} = 0.2\text{ m}\).
• For 0.05 m \(\text{Al}_2(\text{SO}_4)_3\): \(\text{Al}_2(\text{SO}_4)_3\) dissociates into 5 ions (\(2\text{Al}^{3+}\) and \(3\text{SO}_4^{2-}\)). Total concentration of particles = \(5 \times 0.05\text{ m} = 0.25\text{ m}\).
Since 0.05 m \(\text{Al}_2(\text{SO}_4)_3\) produces a higher concentration of particles (0.25 m) than 0.1 m NaCl (0.2 m), it will have a higher freezing point depression. This demonstrates how the van 't Hoff factor directly influences physical properties like freezing point depression.
In simple words: Freezing point depression depends on the total number of dissolved particles. Since Al2(SO4)3 breaks down into 5 ions compared to NaCl's 2 ions, it creates more particles in the solution and lowers the freezing point further.

🎯 Exam Tip: Always calculate the effective molality by multiplying the given concentration by the number of ions (van 't Hoff factor i) to compare colligative properties accurately.

Question viii. State Raoult’s law for a solution containing a nonvolatile solute.
Answer: Statement of Raoult’s law : The law states that the vapour pressure of a solvent over the solution of a nonvolatile solute is equal to the vapour pressure of the pure solvent multiplied by mole fraction of the solvent at constant temperature. This relationship helps in understanding how solutes lower the vapor pressure of a liquid.
In simple words: Raoult's law says that adding a non-volatile substance to a liquid lowers its vapor pressure in proportion to how much substance you add.

🎯 Exam Tip: Always state the condition 'at constant temperature' to secure full marks for this definition.

Question ix. What is the effect on the boiling point of water if 1 mole of methyl alcohol is added to \( 1\text{ dm}^3 \) of water? Why?
Answer:
• The boiling point of water (or any liquid) depends on its vapour pressure.
• Higher the vapour pressure, lower is the boiling point.
• When 1 mole of volatile methyl alcohol is added to \( 1\text{ dm}^3 \) of water, its vapour pressure is increased decreasing the boiling point of water. This occurs because methyl alcohol is highly volatile and escapes into the vapor phase more readily than water.
In simple words: Adding volatile methyl alcohol increases the overall vapor pressure, which makes the mixture boil at a lower temperature than pure water.

🎯 Exam Tip: Remember that adding a volatile solute increases vapor pressure and lowers the boiling point, which is opposite to the effect of a non-volatile solute.

Question x. Which of the four colligative properties is most often used for molecular mass determination? Why?
Answer:
1. Since osmotic pressure has large values, it can be measured more precisely.
2. The osmotic pressure can be measured at a suitable constant temperature.
3. The molecular masses can be measured more accurately. This method is particularly advantageous for biomolecules like proteins which are unstable at high temperatures.
4. Therefore, it is more useful to determine molecular masses of expensive substances by osmotic pressure.
In simple words: Osmotic pressure is preferred because it produces large, easily measurable values at room temperature, making it highly accurate even for delicate or expensive substances.

🎯 Exam Tip: Highlight that osmotic pressure measurements can be carried out at room temperature, which is crucial for biomolecules that degrade at higher temperatures.

Answer the Following

Question i. How vapour pressure lowering is related to a rise in boiling point of solution?
Answer:
(1) The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the external pressure, generally 1 atm (\( 101.3 \times 10^3 \text{ Nm}^{-2} \)).

(2) When a liquid is heated, its vapour pressure rises till it becomes equal to the external pressure. If the liquid has a low vapour pressure, it has a higher boiling point.

Key features of the Vapour Pressure Curve:

• Y-axis represents Vapour pressure (with marked points \( P_0 \) and \( P \))
• X-axis represents Temperature in Kelvin (with marked points \( T_0 \) and \( T \))
• Curve AB represents the pure solvent
• Curve CD represents the solution
• \( \Delta T_b \) represents the elevation in boiling point

(3) When a nonvolatile solute is added to a solvent, its vapour pressure decreases, hence the boiling point increases. This is explained by graphical representation of the vapour pressure and the boiling point of the pure solvent and the solution. If \( T_0 \) and \( T \) are the boiling points of a pure solvent and a solution, then the elevation in the boiling point is given by,
\( \Delta T_b = T - T_0 \)
The curve AB, represents the variation in the vapour pressure of a pure solvent with temperature while curve CD represents the variation in the vapour pressure of the solution.

(4) This elevation in the boiling point is proportional to the lowering of the vapour pressure, i.e., \( P_0 - P \), where \( P_0 \) and \( P \) are the vapour pressures of the pure solvent and the solution respectively. This relationship shows that a greater lowering of vapour pressure always results in a larger elevation of the boiling point.
In simple words: When we add a solute to a liquid, its vapour pressure drops, which means we need to heat it to a higher temperature to make it boil. This difference in boiling temperature is directly related to how much the vapour pressure was lowered.

🎯 Exam Tip: Always draw and label the vapour pressure curves for both the pure solvent and the solution clearly, showing \( \Delta T_b \) as the difference between \( T \) and \( T_0 \) to secure full marks.

Question ii. What are isotonic and hypertonic solutions?
Answer:
(1) Isotonic solutions: The solutions having the same osmotic pressure at a given temperature are called isotonic solutions. This uniform pressure prevents any net movement of solvent between them.

Explanation: If two solutions of substances A and B contain \( n_A \) and \( n_B \) moles dissolved in volume \( V \) (in \( \text{dm}^3 \)) of the solutions, then their concentrations are:
\( C_A = \frac{n_A}{V} \) (in \( \text{mol dm}^{-3} \)) and
\( C_B = \frac{n_B}{V} \) (in \( \text{mol dm}^{-3} \))

If the absolute temperature of both the solutions is \( T \), then by the van’t Hoff equation,
\( \pi_A = C_A RT \) and \( \pi_B = C_B RT \), where \( \pi_A \) and \( \pi_B \) are their osmotic pressures.
For the isotonic solutions,
\( \pi_A = \pi_B \)

\( \implies C_A = C_B \)

\( \implies \frac{n_A}{V} = \frac{n_B}{V} \)

\( \implies n_A = n_B \)
Hence, equal volumes of the isotonic solutions at the same temperature will contain equal number of moles (hence, equal number of molecules) of the substances.

(2) Hypertonic solutions: When two solutions have different osmotic pressures, then the solution having higher osmotic pressure is said to be a hypertonic solution with respect to the other solution.

Explanation: Consider two solutions of substances A and B having osmotic pressures \( \pi_A \) and \( \pi_B \). If \( \pi_B \) is greater than \( \pi_A \), then the solution B is a hypertonic solution with respect to the solution A.
Hence, if \( C_A \) and \( C_B \) are their concentrations, then \( C_B > C_A \). Hence, for equal volume of the solutions, \( n_B > n_A \).
In simple words: Isotonic solutions have the same concentration and osmotic pressure, so water doesn't flow between them. Hypertonic solutions have a higher concentration and osmotic pressure compared to another solution.

🎯 Exam Tip: Clearly define both terms and show the mathematical derivation using the van't Hoff equation to secure full marks. Remember that isotonic solutions must have equal molar concentrations at the same temperature.

Question iii. A solvent and its solution containing a nonvolatile solute are separated by a
Answer: semipermeable membrane. This membrane allows only solvent molecules to pass through while blocking the larger solute particles.
In simple words: A semipermeable membrane acts like a tiny filter that lets water pass through but stops the dissolved solute.

🎯 Exam Tip: Always use the term 'semipermeable membrane' when describing the separation of solvent and solution in osmosis questions.

Question. Does the flow of solvent occur in both directions through a semipermeable membrane? Comment giving reason.
Answer: Yes, the flow of solvent occurs in both directions.
1. When a solvent and a solution containing a non-volatile solute are separated by a semipermeable membrane, there arises a flow of solvent molecules from solvent to solution as well as from solution to solvent.
2. Due to higher vapour pressure of solvent than solution, the rate of flow of solvent molecules from solvent to solution is higher.
3. As more and more solvent passes into solution due to osmosis, the solvent content increases, and the rate of backward flow increases.
4. At a certain stage an equilibrium is reached where both the opposing rates become equal attaining an equilibrium.
In simple words: Solvent molecules actually move both ways across the membrane, but they flow faster from the pure solvent side to the solution side until the system reaches a balance.

🎯 Exam Tip: Clearly state that flow occurs in both directions, but emphasize that the net flow is from solvent to solution until equilibrium is reached.

Question iv. The osmotic pressure of \( \text{CaCl}_2 \) and urea solutions of the same concentration at the same temperature are respectively 0.605 atm and 0.245 atm. Calculate van’t Hoff factor for \( \text{CaCl}_2 \).
Answer:
Given:
\( \pi_{\text{CaCl}_2} = 0.605 \text{ atm} \)
\( \pi_{\text{urea}} = 0.245 \text{ atm} \)
For urea solution, van’t Hoff factor, \( i = 1 \)
\( \pi_{\text{CaCl}_2} = i \times (CRT)_{\text{CaCl}_2} \)
\( \pi_{\text{urea}} = (CRT)_{\text{urea}} \)
Since concentration (C) and temperature (T) are the same:
\( \frac{\pi_{\text{CaCl}_2}}{\pi_{\text{urea}}} = \frac{i(CRT)}{(CRT)} \)
\( \implies i = \frac{\pi_{\text{CaCl}_2}}{\pi_{\text{urea}}} \)
\( \implies i = \frac{0.605}{0.245} = 2.47 \)
Thus, the van’t Hoff factor for \( \text{CaCl}_2 \) is 2.47.
In simple words: By comparing the osmotic pressure of calcium chloride to urea (which doesn't split into ions), we find that calcium chloride produces 2.47 times more particles in the solution.

🎯 Exam Tip: Remember that urea is a non-electrolyte so its van't Hoff factor \( i \) is always 1. Show the formula ratio clearly to secure full marks.

Question v. Explain reverse osmosis.
Answer: Reverse osmosis: The phenomenon of the passage of solvent like water under high pressure from the concentrated aqueous solution like seawater into pure water through a semipermeable membrane is called reverse osmosis. This process is highly effective for desalination and is widely used globally to produce clean drinking water.
In simple words: Reverse osmosis is when we apply high pressure to salty or dirty water to force pure water backward through a special filter, leaving all the impurities behind.

🎯 Exam Tip: Always mention that the applied pressure must be greater than the osmotic pressure for reverse osmosis to occur.

The osmotic pressure of seawater is about 30 atmospheres. Hence when pressure more than 30 atmospheres is applied on the solution side, regular osmosis stops and reverse osmosis starts. Hence pure water from seawater enters the other side of pure water.

Purification of Seawater by Reverse Osmosis

For this purpose of suitable semipermeable membrane is required which can withstand high pressure conditions over a long period. This method is used successfully in Florida since 1981 producing more than 10 million litres of pure water per day.

Question vi. How molar mass of a solute is determined by osmotic pressure measurement?
Answer: Consider \( V \text{ dm}^3 \) (litres) of a solution containing \( W_2 \) mass of a solute of molar mass \( M_2 \) at a temperature \( T \).
Number of moles of solute, \( n_2 = \frac{W_2}{M_2} \)
The osmotic pressure \( \pi \) is given by,
\( \pi = \frac{W_2RT}{M_2V} \)
\( \therefore M_2 = \frac{W_2RT}{\pi V} \)
By measuring osmotic pressure of a solution, the molar mass of a solute can be calculated. Since osmotic pressure can be measured more precisely than other colligative properties, it is widely used to measure molar masses of the substances, especially for macromolecules like proteins and polymers.
In simple words: We can find the weight of a tiny molecule by measuring how much pressure its solution creates. By putting the measured pressure and temperature into a simple formula, we can easily calculate the molecule's molar mass.

🎯 Exam Tip: Always write down the final formula \( M_2 = \frac{W_2RT}{\pi V} \) clearly and define each term with its correct unit to secure full marks.

Question vii. Why vapour pressure of a solvent is lowered by dissolving a nonvolatile solute into it?
Answer: When a nonvolatile solute is dissolved in a pure solvent, some of the solvent molecules at the surface are replaced by nonvolatile solute molecules. Since these solute molecules do not escape into the vapour phase, the rate of evaporation of the solvent decreases. Consequently, fewer solvent molecules escape into the gas phase, which directly results in a lower vapour pressure compared to the pure solvent.
In simple words: When you add solute particles, they block some of the surface area of the liquid. This makes it harder for the liquid molecules to escape into the air, which lowers the overall vapour pressure.

🎯 Exam Tip: Mention that surface area is occupied by nonvolatile solute particles to make your explanation highly effective for examiners.

Question viii. Using Raoult’s law, how will you show that \( \Delta P = P^0_1 x_2 \)? Where \( x_2 \) is the mole fraction of solute in the solution and \( P^0_1 \) vapour pressure of pure solvent.
Answer: Let \( x_1 \) and \( x_2 \) be the mole fractions of solvent and solute respectively. We know that: \( x_1 + x_2 = 1 \)
\( \implies x_1 = 1 - x_2 \) According to Raoult's law, the vapour pressure of a solution containing a non-volatile solute (\( P \)) is given by: \( P = x_1 P^0_1 \) where \( P^0_1 \) is the vapour pressure of the pure solvent. Substituting the value of \( x_1 \): \( P = (1 - x_2) P^0_1 \)
\( \implies P = P^0_1 - P^0_1 x_2 \)
\( \implies P^0_1 x_2 = P^0_1 - P \) Since the lowering of vapour pressure is defined as \( \Delta P = P^0_1 - P \), we can substitute this into the equation:
\( \implies \Delta P = P^0_1 x_2 \) This successfully demonstrates that the lowering of vapour pressure is directly proportional to the mole fraction of the solute.
In simple words: Raoult's law shows that when you add a solute to a solvent, the drop in vapour pressure depends directly on the fraction of solute particles present in the mixture.

🎯 Exam Tip: Clearly define each term like \( x_1 \), \( x_2 \), and \( P^0_1 \) at the beginning of the derivation to ensure you get full marks for step-by-step presentation.

Question ix. While considering boiling point elevation and freezing point depression a solution concentration is expressed in molality and not in molarity. Why?
Answer: Boiling point elevation and freezing point depression involve temperature changes (\( \Delta T_b \) and \( \Delta T_f \)). Molarity depends on the volume of the solution, which changes as temperature changes. In contrast, molality is based on the mass of the solvent, which remains completely unaffected by temperature variations. Therefore, to keep concentration values stable across different temperatures, we express concentration in molality rather than molarity.
In simple words: Molarity changes with temperature because liquids expand or contract when heated. Molality is based on mass, which stays exactly the same no matter how hot or cold it gets.

🎯 Exam Tip: Always highlight that volume is temperature-dependent while mass is temperature-independent to secure full marks on this classic conceptual question.

Question 4. Derive the relationship between degree of dissociation of an electrolyte and van’t Hoff factor.
Answer: Consider \( 1\text{ dm}^3 \) of a solution containing \( m \) moles of an electrolyte \( A_xB_y \). The electrolyte on dissociation gives \( x \) number of \( A^{y+} \) ions and \( y \) number of \( B^{x-} \) ions. Let \( \alpha \) be the degree of dissociation.

At equilibrium,
\( A_xB_y \rightleftharpoons xA^{y+} + yB^{x-} \)

For 1 mole of electrolyte: \( 1 - \alpha \), \( x\alpha \), \( y\alpha \) and
For \( m \) moles of an electrolyte: \( m(1 - \alpha) \), \( mx\alpha \), \( my\alpha \) are the number of particles.

Total number of moles at equilibrium will be:
Total moles \( = m(1 - \alpha) + mx\alpha + my\alpha \)
\( = m[(1 - \alpha) + x\alpha + y\alpha] \)
\( = m[1 - \alpha + x\alpha + y\alpha] \)
\( = m[1 + \alpha(x + y - 1)] \)

The van’t Hoff factor \( i \) will be:
\( i = \frac{\text{Observed colligative property}}{\text{Theoretical colligative property}} \)

\( i = \frac{m[1 + \alpha(x + y - 1)]}{m} \)

\( i = 1 + \alpha(x + y - 1) \)

If the total number of ions from one mole of electrolyte is denoted by \( n \), then \( (x + y) = n \).
\( \therefore i = 1 + \alpha(n - 1) \)
\( \therefore \alpha(n - 1) = i - 1 \)
\( \therefore \alpha = \frac{i - 1}{n - 1} \) .......(1)

This is the relation between van’t Hoff factor \( i \) and degree of dissociation of an electrolyte.
In simple words: This formula helps us find how much a chemical splits into smaller pieces when dissolved in water by comparing the actual number of particles to what we started with.

🎯 Exam Tip: Clearly define each term like \( \alpha \), \( i \), and \( n \) at the beginning of the derivation to secure full marks.

Question 5. What is effect of temperature on solubility of solids in water? Give examples.
Answer: Generally, the solubility of solid solutes in water increases with an increase in temperature. For example, the solubility of substances like \( \text{KCl} \), \( \text{KNO}_3 \), and \( \text{NaCl} \) increases when temperature rises. However, for some substances like \( \text{Li}_2\text{SO}_4 \text{ and } \text{Ce}_2(\text{SO}_4)_3 \), the solubility decreases with an increase in temperature because their dissolution process is exothermic.
In simple words: For most solids, heating the water helps more of the solid dissolve, like sugar in hot tea. But for a few special solids, heating actually makes them dissolve less.

🎯 Exam Tip: Always mention both cases (solubility increasing and decreasing with temperature) along with at least one example for each to get full marks.

Solubility of a Solid Solute

The solubility of a solid solute depends upon temperature.

Variation of Solubilities of Some Ionic Solids with Temperature

• Generally rise in temperature increases the solubility. This is due to expansion of holes or empty spaces in the liquid solvent. Generally 10 °C rise in temperature, increases the solubility of solids two fold.
• Dissolution process may be endothermic or exothermic.
• The solubility of the substances like NaBr, NaCl, KCl, etc. changes slightly with the increase in temperature.
• The solubility of the salts like \( \text{NaNO}_3 \), \( \text{KNO}_3 \), KBr, etc. increases appreciably with the increase in temperature.
• The solubility of \( \text{Na}_2\text{SO}_4 \) first increases and after 30 °C decreases with the increase in temperature.

This variation in solubility with temperature can be used to separate the salts from the mixture by fractional crystallisation.

Question 6. Obtain the relationship between freezing point depression of a solution containing nonvolatile nonelectrolyte and its molar mass.
Answer: The freezing point depression, \( \Delta T_f \) of a solution is directly proportional to molality (m) of the solution.
\( \implies \Delta T_f \propto m \)
\( \implies \Delta T_f = K_f m \)
In simple words: The drop in freezing point of a liquid when you add a solute depends directly on how many solute particles are dissolved in it.

🎯 Exam Tip: Always state the proportionality relationship first before introducing the constant \( K_f \) to show a logical step-by-step derivation.

where \( K_f \) is a molal depression constant. The molality of a solution is given by, \[ m = \frac{\text{Number of moles of the solute}}{\text{Weight of the solvent in kg}} \] If \( W_1 \) grams of a solvent contain \( W_2 \) grams of a solute of the molar mass \( M_2 \), then the molality \( m \) of the solution is given by, \[ m = \frac{W_2 \times 1000}{W_1 \times M_2} \text{ mol kg}^{-1} \] \[ \therefore \Delta T_f = K_f \times \frac{W_2 \times 1000}{W_1 M_2} \] If the weights are expressed in kg then, \[ \Delta T_f = K_f \times \frac{W_2}{W_1 M_2} \] The unit of \( K_f \) is \( \text{K kg mol}^{-1} \) Hence, from the measurement of the depression in the freezing point of the solution, the molar mass of the substance can be determined.

Question 7. Explain with diagram the boiling point elevation in terms of vapour pressure lowering.
Answer:
(1) The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the external pressure, generally 1 atm \( (101.3 \times 10^3 \text{ Nm}^{-2}) \). Understanding this relationship is crucial for studying colligative properties.
(2) When a liquid is heated, its vapour pressure rises till it becomes equal to the external pressure. If the liquid has a low vapour pressure, it has a higher boiling point.
In simple words: The boiling point is the temperature where a liquid's vapour pressure equals the surrounding atmospheric pressure. If a solution has a lower vapour pressure than the pure solvent, it needs to be heated to a higher temperature to boil.

🎯 Exam Tip: Always define boiling point in terms of vapour pressure matching external pressure (1 atm) to secure full marks, and clearly state the inverse relationship between vapour pressure and boiling point.

Vapour Pressure Curve Showing Elevation in Boiling Point

Key elements of the vapour pressure curve diagram:

• Y-axis: Vapour pressure (from P to 1.0 atm)
• X-axis: Temperature in Kelvin (K) (from \( T_0 \) to \( T \))
• Curve AB: Represents the vapour pressure of the pure solvent
• Curve CD: Represents the vapour pressure of the solution
• \( \Delta T_b \): Elevation in boiling point (\( T - T_0 \))

(3) When a nonvolatile solute is added to a solvent, its vapour pressure decreases, hence the boiling point increases. This is explained by graphical representation of the vapour pressure and the boiling point of the pure solvent and the solution.

If \( T_0 \) and \( T \) are the boiling points of a pure solvent and a solution, then the elevation in the boiling point is given by:
\( \Delta T_b = T - T_0 \)

The curve AB represents the variation in the vapour pressure of a pure solvent with temperature, while curve CD represents the variation in the vapour pressure of the solution.

(4) This elevation in the boiling point is proportional to the lowering of the vapour pressure, i.e., \( P_0 - P \), where \( P_0 \) and \( P \) are the vapour pressures of the pure solvent and the solution.
\( [ \Delta T_b \propto (P_0 - P) \text{ or } \Delta T_b \propto \Delta P ] \)

Question 8. Fish generally needs \( \text{O}_2 \) concentration in water at least 3.8 mg/L for survival. What partial pressure of \( \text{O}_2 \) above the water is needed for the survival of fish? Given the solubility of \( \text{O}_2 \) in water at 0 °C and 1 atm partial pressure is \( 2.2 \times 10^{-3} \text{ mol/L} \) (0.054 atm)
Answer:
Given: Required concentration of \( \text{O}_2 \) (\( S \)) = 3.8 mg/L = \( \frac{3.8 \times 10^{-3}}{32} \text{ mol L}^{-1} = 1.187 \times 10^{-4} \text{ mol L}^{-1} \)
Solubility of \( \text{O}_2 \) (\( S_1 \)) at \( P_1 = 1 \text{ atm} \) is \( 2.2 \times 10^{-3} \text{ mol L}^{-1} \).
According to Henry's law:
\( S = K_H \times P \)
\( \implies K_H = \frac{S_1}{P_1} = \frac{2.2 \times 10^{-3} \text{ mol L}^{-1}}{1 \text{ atm}} = 2.2 \times 10^{-3} \text{ mol L}^{-1} \text{ atm}^{-1} \)
To find the required partial pressure (\( P_2 \)) for survival:
\( S = K_H \times P_2 \)
\( \implies P_2 = \frac{S}{K_H} = \frac{1.187 \times 10^{-4} \text{ mol L}^{-1}}{2.2 \times 10^{-3} \text{ mol L}^{-1} \text{ atm}^{-1}} \)
\( \implies P_2 = 0.054 \text{ atm} \)
Thus, the partial pressure of \( \text{O}_2 \) needed above the water for the survival of fish is 0.054 atm.
In simple words: To survive, fish need a minimum amount of oxygen dissolved in water. Using Henry's Law, we calculate that a partial pressure of 0.054 atm of oxygen is required to maintain this necessary concentration.

🎯 Exam Tip: Always convert the solubility concentration from mg/L to mol/L by dividing by the molar mass of oxygen (32 g/mol) before substituting it into Henry's Law formula.

Question 9. The vapour pressure of water at 20 °C is 17 mm Hg. What is the vapour pressure of solution containing 2.8 g urea in 50 g of water? (16.17 mm Hg)
Answer:
Given:
Vapour pressure of pure solvent (water), \( P^0 = 17 \text{ mm Hg} \)
Weight of solvent, \( W_1 = 50 \text{ g} \)
Weight of solute (urea), \( W_2 = 2.8 \text{ g} \)
Molecular weight of solvent, \( M_1 = 18 \text{ g/mol} \)
Molecular weight of solute (urea), \( M_2 = 60 \text{ g/mol} \)

According to Raoult's law for relative lowering of vapour pressure:
\( \frac{P^0 - P}{P^0} = \frac{W_2 \times M_1}{W_1 \times M_2} \)

Substituting the values:
\( \frac{17 - P}{17} = \frac{2.8 \times 18}{50 \times 60} \)

\( \frac{17 - P}{17} = \frac{50.4}{3000} \)

\( \frac{17 - P}{17} = 0.0168 \)

\( \implies 17 - P = 17 \times 0.0168 \)

\( \implies 17 - P = 0.2856 \)

\( \implies P = 17 - 0.2856 \)

\( \implies P = 16.7144 \text{ mm Hg} \)

The vapour pressure of the solution is \( 16.7144 \text{ mm Hg} \). This calculation is based on Raoult's law for non-volatile solutes, which relates the relative lowering of vapour pressure to the mole fraction of the solute.
In simple words: When a non-volatile substance like urea is dissolved in water, it makes it harder for water molecules to escape into the air, which lowers the overall vapour pressure. By using the formula for relative lowering of vapour pressure, we find that the new vapour pressure of the solution is 16.7144 mm Hg.

🎯 Exam Tip: Always state the formula clearly before substituting values, and remember to write the final unit (mm Hg) to secure full marks.

Question 10. A 5% aqueous solution (by mass) of cane sugar (molar mass 342 g/mol) has freezing point of 271K. Calculate the freezing point of 5% aqueous glucose solution.
Answer:
1. For 5% cane sugar solution:
Mass of cane sugar (\( W_2 \)) = 5 g
Mass of solvent water (\( W_1 \)) = 100 g - 5 g = 95 g
Molar mass of cane sugar (\( M_2 \)) = 342 g/mol
Freezing point of pure water (\( T_f^0 \)) = 273.15 K
Freezing point of cane sugar solution (\( T_f \)) = 271 K
Depression in freezing point (\( \Delta T_f \)) = \( T_f^0 - T_f = 273.15 \text{ K} - 271 \text{ K} = 2.15 \text{ K} \)

We know that:
\( \Delta T_f = \frac{K_f \times W_2 \times 1000}{M_2 \times W_1} \)

For cane sugar:
\( 2.15 = \frac{K_f \times 5 \times 1000}{342 \times 95} \) ——— (Equation 1)

2. For 5% glucose solution:
Mass of glucose (\( W_2' \)) = 5 g
Mass of solvent water (\( W_1' \)) = 95 g
Molar mass of glucose (\( M_2' \)) = 180 g/mol

For glucose:
\( \Delta T_f' = \frac{K_f \times 5 \times 1000}{180 \times 95} \) ——— (Equation 2)

Dividing Equation 2 by Equation 1:
\( \frac{\Delta T_f'}{2.15} = \frac{342}{180} \)

\( \implies \frac{\Delta T_f'}{2.15} = 1.9 \)

\( \implies \Delta T_f' = 1.9 \times 2.15 = 4.085 \text{ K} \)

Now, the freezing point of the glucose solution (\( T_f' \)) is:
\( T_f' = T_f^0 - \Delta T_f' \)

\( \implies T_f' = 273.15 \text{ K} - 4.085 \text{ K} = 269.065 \text{ K} \)

Therefore, the freezing point of the 5% aqueous glucose solution is \( 269.065 \text{ K} \). This shows how a lower molar mass solute causes a greater depression in freezing point for the same mass percentage.
In simple words: Since glucose molecules are lighter than cane sugar molecules, a 5% glucose solution contains more solute particles than a 5% sugar solution. More particles mean a greater drop in the freezing point, making the glucose solution freeze at a lower temperature of 269.065 K.

🎯 Exam Tip: When comparing two solutions, dividing their freezing point depression equations is the fastest way to eliminate the constant \( K_f \) and avoid tedious calculations.

Question 11. A solution of citric acid \( \text{C}_6\text{H}_8\text{O}_7 \) in \( 50 \text{ g} \) of acetic acid has a boiling point elevation of \( 1.76 \text{ K} \). If \( K_b \) for acetic acid is \( 3.07 \text{ K kg mol}^{-1} \), what is the molality of solution?
Answer:
Given:
Mass of solvent (acetic acid), \( W_1 = 50 \text{ g} \)
Elevation of boiling point, \( \Delta T_b = 1.76 \text{ K} \)
Boiling point elevation constant, \( K_b = 3.07 \text{ K kg mol}^{-1} \)
Molality of solution, \( m = ? \)

We know that,
\( \Delta T_b = K_b \times m \)

Rearranging the formula to solve for molality:
\( \implies m = \frac{\Delta T_b}{K_b} \)
\( \implies m = \frac{1.76}{3.07} \)
\( \implies m = 0.5733 \text{ m} \)

Thus, the molality of the citric acid solution is \( 0.5733 \text{ m} \). This concentration indicates the number of moles of citric acid dissolved per kilogram of acetic acid solvent.
In simple words: Molality tells us how crowded the solute particles are in a solvent. By dividing the temperature change by the constant value for acetic acid, we find that there are about 0.5733 moles of citric acid in every kilogram of acetic acid.

🎯 Exam Tip: Always write down the given values with their correct units first, as this helps you select the right formula and prevents calculation errors.

Question 13. A mixture of benzene and toluene contains 30% by mass of toluene. At 30°C, vapour pressure of pure toluene is 36.7 mm Hg and that of pure benzene is 118.2 mm Hg. Assuming that the two liquids form ideal solutions, calculate the total pressure and partial pressure of each constituent above the solution at 30°C.
Answer:
Given: 30% by mass of toluene (T) and 70% by mass of benzene (B).
\( W_T = 30 \text{ g} \); \( W_B = 70 \text{ g} \)
\( P^0_T = 36.7 \text{ mm Hg} \); \( P^0_B = 118.2 \text{ mm Hg} \)
\( M_T = 92 \text{ g mol}^{-1} \); \( M_B = 78 \text{ g mol}^{-1} \)
\( P_T = ? \), \( P_B = ? \), \( P_{\text{soln}} = ? \)

Number of moles:
\( n_T = \frac{W_T}{M_T} = \frac{30}{92} = 0.3260 \text{ mol} \)
\( n_B = \frac{W_B}{M_B} = \frac{70}{78} = 0.8974 \text{ mol} \)

Total number of moles:
\( n_{\text{Total}} = n_T + n_B \)
\( = 0.326 + 0.8974 \)
\( = 1.2234 \text{ mol} \)

Mole fractions:
\( x_T = \frac{n_T}{n_{\text{Total}}} = \frac{0.326}{1.2234} = 0.2665 \)
\( x_B = 1 - 0.2665 = 0.7335 \)

Total pressure of solution:
\( P_{\text{soln}} = (x_T \times P^0_T) + (x_B \times P^0_B) \)
\( = (0.2665 \times 36.7) + (0.7335 \times 118.2) \)
\( = 9.780 + 86.7 \)
\( = 96.48 \text{ mm Hg} \)

Partial pressures:
\( P_T = x_T \times P_{\text{soln}} \)
\( = 0.2665 \times 96.48 \)
\( = 25.71 \text{ mm Hg} \)
\( P_B = x_B \times P_{\text{soln}} \)
\( = 0.7335 \times 96.48 \)
\( = 70.77 \text{ mm Hg} \)

Total pressure \( P_{\text{soln}} = 96.48 \text{ mm Hg} \). This calculation confirms that the total vapor pressure is the sum of the individual partial vapor pressures of the components.
In simple words: To find the pressures, we first calculate how many moles of toluene and benzene we have based on their mass. Then, we find their mole fractions and multiply them by their individual pure vapor pressures to get the partial and total pressures of the mixture.

🎯 Exam Tip: Always write down the formula for Raoult's law clearly before substituting values, as step-wise marks are awarded for formulas and correct units like mm Hg.

Question 14. At 25 °C a 0.1 molal solution of CH₃COOH is 1.35% dissociated in an aqueous solution. Calculate freezing point and osmotic pressure of the solution assuming molality and molarity to be identical.
Answer:
Given:
\( T = 273 + 25 = 298 \text{ K} \)
\( C = 0.1 \text{ m} \cong 0.1 \text{ M} \)
\( K_f = 1.86 \text{ K kg mol}^{-1} \)
Per cent dissociation = \( 1.35 \)
Freezing point \( = t_f = ? \)
\( \pi = ? \)
\( \alpha = \frac{1.35}{100} = 0.0135 \)

\( \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \)
\( \begin{matrix} 1-\alpha & \quad\quad\quad\quad \alpha & \quad\quad \alpha \end{matrix} \)

\( i = 1 - \alpha + \alpha + \alpha = 1 + \alpha = 1 + 0.0135 = 1.0135 \)

(i) \( \Delta T_f = i \times K_f \times m \)
\( = 1.0135 \times 1.86 \times 0.1 \)
\( = 0.1885 \text{ }^\circ\text{C} \)
\( \therefore \text{Freezing point of solution} = 0 - 0.1885 \)
\( = -0.1885 \text{ }^\circ\text{C} \)

(ii) \( \pi = iCRT \)
\( = 1.035 \times 0.1 \times 0.08206 \times 298 \)
\( = 2.53 \text{ atm} \)

(i) Freezing point of solution \( = -0.1885 \text{ }^\circ\text{C} \)
(ii) Osmotic pressure \( = \pi = 2.53 \text{ atm} \)
In simple words: We first find the van 't Hoff factor (i) using the percentage of dissociation. Then, we use this factor to calculate how much the freezing point drops and what the osmotic pressure of the solution will be.

🎯 Exam Tip: Always remember to convert the percentage of dissociation into a fraction (\( \alpha \)) before using it in the formula for the van 't Hoff factor \( i \).

Question 15. A 0.15 m aqueous solution of KCl freezes at -0.510 °C. Calculate i and osmotic pressure at 0 °C. Assume volume of solution equal to that of water.
Answer:
Given:
\( c = 0.15 \text{ m KCl} \cong 0.15 \text{ M KCl} \)
\( \Delta T_f = 0 - T_f = 0 - (-0.510) = 0.510 \text{ }^\circ\text{C} \)
This depression in freezing point will help us determine the van 't Hoff factor for the solute.
In simple words: We start by calculating the depression in freezing point (\( \Delta T_f \)) from the given freezing point of the solution.

🎯 Exam Tip: Remember that the freezing point of pure water is \( 0\text{ }^\circ\text{C} \), so the depression in freezing point \( \Delta T_f \) is always \( 0 - T_f \).

Can You Tell? (Textbook Page No. 29)

Question 1. Why naphthalene dissolves in benzene but not in water?
Answer: Since naphthalene is a covalent nonpolar substance, it is soluble in a nonpolar solvent like benzene but insoluble in polar solvent like water. This behavior is governed by the fundamental chemical principle of 'like dissolves like'.
In simple words: Naphthalene and benzene are both nonpolar substances, so they mix together easily. Water is polar, which is why nonpolar naphthalene cannot dissolve in it.

🎯 Exam Tip: To get full marks, always explicitly mention the terms 'polar' and 'nonpolar' for both the solute and the solvent.

Question 2. Anhydrous sodium sulphate dissolves in water with the evolution of heat. What is the effect of temperature on its solubility?
Answer: Since the dissolution of anhydrous sodium sulphate in water is an exothermic process due to evolution of heat, according to Le Chatelier’s principle its solubility decreases with the increase in temperature. This means that adding external heat shifts the equilibrium back towards the undissolved state.
In simple words: When sodium sulphate dissolves, it naturally releases heat. If we heat the water even more, it opposes this process and makes the substance less soluble.

🎯 Exam Tip: Always state Le Chatelier's principle clearly when explaining the effect of temperature on solubility for exothermic or endothermic processes.

Textbook Page No. 42

Question 1. If \( 1.25\text{ m} \) sucrose solution has \( \Delta T_f \) of \( 2.32\text{ }^\circ\text{C} \), what will be the expected value of \( \Delta T_f \) for \( 1.25\text{ m CaCl}_2 \) solution?
Answer: Sucrose being nonelectrolyte, it has \( i = 1 \) but for \( \text{CaCl}_2 \), (\( \text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2\text{Cl}^- \)) the value of \( i = 3 \). Hence, the expected value of \( \Delta T_f \) for \( 1.25\text{ m CaCl}_2 \) solution is three times that of sucrose, which is \( 3 \times 2.32\text{ }^\circ\text{C} = 6.96\text{ }^\circ\text{C} \). This significant increase occurs because the electrolyte dissociates completely into three active ions in the solution.
In simple words: Sucrose does not split into pieces when dissolved, but calcium chloride splits into three ions. Because there are three times as many particles, the freezing point drops three times as much.

🎯 Exam Tip: Show the dissociation equation of the electrolyte to clearly justify why the van 't Hoff factor \( i \) is equal to 3.

Question 1. Which of the following solutions will have maximum boiling point elevation and which have minimum freezing point depression assuming the complete dissociation? (a) 0.1m KCl (b) 0.05 m NaCl (c) 1 m \( \text{AlPO}_4 \) (d) 0.1 m \( \text{MgSO}_4 \).
Answer: Boiling point elevation and freezing point depression are colligative properties that depend on the number of particles in solution. The solution having more number of particles will have large boiling point elevation and that having less number of particles would show minimum freezing point depression. This relationship highlights how colligative properties depend solely on the concentration of solute particles rather than their chemical identity.
(a) \( \text{KCl} \rightarrow \text{K}^+ + \text{Cl}^- \) (Total particles in solution = 0.2 mol)
(b) \( \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \) (Total particles in solution = 0.1 mol)
(c) \( \text{AlPO}_4 \rightarrow \text{Al}^{3+} + \text{PO}_4^{3-} \) (Total particles in solution = 2.0 mol)
(d) \( \text{MgSO}_4 \rightarrow \text{Mg}^{2+} + \text{SO}_4^{2-} \) (Total particles in solution = 0.2 mol)
\( \text{AlPO}_4 \) solution contains the highest moles and hence the highest number of particles and in turn, the maximum \( \Delta T_b \). \( \text{NaCl} \) solution has minimum moles and particles, so it has minimum \( \Delta T_f \).
In simple words: Boiling point elevation and freezing point depression depend on the total number of dissolved particles. Since \( \text{AlPO}_4 \) breaks down into the most particles, it raises the boiling point the most, while \( \text{NaCl} \) has the fewest particles and lowers the freezing point the least.

🎯 Exam Tip: Always write down the dissociation equations and calculate the total concentration of ions (\( i \times m \)) to justify your answer clearly to the examiner.

Question 2. Arrange the following solutions in order of increasing osmotic pressure. Assume complete ionization. (a) 0.5 m \( \text{Li}_2\text{SO}_4 \) (b) 0.5 m KCl (c) 0.5 m \( \text{Al}_2(\text{SO}_4)_3 \) (d) 0.1 m \( \text{BaCl}_2 \).
Answer: Consider \( 1\text{ dm}^3 \) of each solution. Osmotic pressure (\( \pi \)) is a colligative property and depends on the concentration of solute particles (ions) in the solution. This order directly reflects how the total number of active particles in solution dictates the magnitude of osmotic pressure.
The concentration of particles for each solution is calculated as follows:
(a) \( 0.5\text{ m Li}_2\text{SO}_4 \rightarrow 2\text{Li}^+ + \text{SO}_4^{2-} \) (3 ions)
\( \implies \text{Concentration of particles} = 3 \times 0.5\text{ m} = 1.5\text{ m} \)
(b) \( 0.5\text{ m KCl} \rightarrow \text{K}^+ + \text{Cl}^- \) (2 ions)
\( \implies \text{Concentration of particles} = 2 \times 0.5\text{ m} = 1.0\text{ m} \)
(c) \( 0.5\text{ m Al}_2(\text{SO}_4)_3 \rightarrow 2\text{Al}^{3+} + 3\text{SO}_4^{2-} \) (5 ions)
\( \implies \text{Concentration of particles} = 5 \times 0.5\text{ m} = 2.5\text{ m} \)
(d) \( 0.1\text{ m BaCl}_2 \rightarrow \text{Ba}^{2+} + 2\text{Cl}^- \) (3 ions)
\( \implies \text{Concentration of particles} = 3 \times 0.1\text{ m} = 0.3\text{ m} \)
Therefore, the order of increasing osmotic pressure is:
\( 0.1\text{ m BaCl}_2 < 0.5\text{ m KCl} < 0.5\text{ m Li}_2\text{SO}_4 < 0.5\text{ m Al}_2(\text{SO}_4)_3 \)
i.e., (d) < (b) < (a) < (c).
In simple words: Osmotic pressure increases when there are more dissolved particles in the solution. By multiplying the concentration of each solution by the number of ions it splits into, we can easily arrange them from lowest to highest concentration of particles.

🎯 Exam Tip: Remember that osmotic pressure is directly proportional to the van 't Hoff factor (\( i \)) multiplied by the concentration (\( M \) or \( m \)). Always show this calculation step-by-step to secure full marks.

Question 1. Arrange the following solutions in the increasing order of their osmotic pressure: \( 0.5\text{ m Li}_2\text{SO}_4 \), \( 0.5\text{ m KCl} \), \( 0.5\text{ m Al}_2(\text{SO}_4)_3 \), and \( 0.1\text{ m BaCl}_2 \).
Answer: To find the increasing order of osmotic pressure, we must calculate the total concentration of particles (ions) after dissociation for each solution:

(a) \( 0.5\text{ m Li}_2\text{SO}_4 \rightarrow 2\text{Li}^+ + \text{SO}_4^{2-} \)
Total particles = \( (2 \times 0.5\text{ m}) + 0.5\text{ m} = 1.5\text{ m} \)

(b) \( 0.5\text{ m KCl} \rightarrow \text{K}^+ + \text{Cl}^- \)
Total particles = \( 0.5\text{ m} + 0.5\text{ m} = 1\text{ m} \)

(c) \( 0.5\text{ m Al}_2(\text{SO}_4)_3 \rightarrow 2\text{Al}^{3+} + 3\text{SO}_4^{2-} \)
Total particles = \( (2 \times 0.5\text{ m}) + (3 \times 0.5\text{ m}) = 1 + 1.5 = 2.5\text{ m} \)

(d) \( 0.1\text{ m BaCl}_2 \rightarrow \text{Ba}^{2+} + 2\text{Cl}^- \)
Total particles = \( 0.1\text{ m} + (2 \times 0.1\text{ m}) = 0.3\text{ m} \)

Osmotic pressure being a colligative property, it depends on the number of particles in the solution. This relationship shows how ionic dissociation directly influences physical properties like osmotic pressure.
Therefore, the increasing order of osmotic pressure is:
\( \text{BaCl}_2\ (0.3\text{ m}) < \text{KCl}\ (1\text{ m}) < \text{Li}_2\text{SO}_4\ (1.5\text{ m}) < \text{Al}_2(\text{SO}_4)_3\ (2.5\text{ m}) \)
In simple words: Osmotic pressure depends on the total number of dissolved particles in a solution. By calculating how each compound breaks down into ions, we can find which solution has the fewest particles and which has the most to arrange them in order.

🎯 Exam Tip: Always write down the dissociation equation for each electrolyte to calculate the total concentration of ions accurately before comparing colligative properties.