CUET UG Chemistry Booster Test - 2 Measurement and Applications of Conductivity
📌 Answers are locked once submitted — results and explanations appear at the end.
QUESTION 1 OF 20
Identify the relationship used to determine the cell constant (G*) using standard KCl solutions.
QUESTION 2 OF 20
Why is the measurement of l and A to calculate the cell constant considered inconvenient and unreliable?
QUESTION 3 OF 20
Identify the correct statements regarding the Wheatstone bridge method for measuring resistance of a conductivity cell.
Statements:
1. The bridge is balanced when no current passes through the detector.
2. Direct current (DC) is used to prevent polarization.
3. An audio-frequency oscillator is used as the AC source.
4. The unknown resistance R₂ is calculated as (R₁R₄)/R₃.
QUESTION 4 OF 20
When a Wheatstone bridge is balanced, the measured resistance is in ohms (Ω). The unit of conductance (G), which is the inverse of resistance, is:
QUESTION 5 OF 20
Arrange the following in decreasing order of their typical conductivity at 298.15 K:
1. Copper
2. 0.1 M HCl(aq)
3. Teflon
4. Pure water
QUESTION 6 OF 20
Match List-I (Quantity) with List-II (SI Unit).
| List I | List II |
|---|---|
| 1. Resistivity (ρ) | b. Ω m |
| 2. Cell constant (G*) | d. m⁻¹ |
| 3. Conductivity (κ) | a. S m⁻¹ |
| 4. Molar conductivity (Λm) | c. S m² mol⁻¹ A.1-d, 2-a, 3-c, 4-b |
QUESTION 7 OF 20
The molar conductivity of a solution increases with a decrease in concentration because:
QUESTION 8 OF 20
If conductivity (κ) is expressed in S cm⁻¹ and concentration (c) in mol L⁻¹, the formula for molar conductivity (Λm) in S cm² mol⁻¹ is:
QUESTION 9 OF 20
According to the passage, why does conductivity always decrease with dilution for all electrolytes?
QUESTION 10 OF 20
According to the passage, conductivity is equivalent to the conductance of what specific volume of solution?
QUESTION 11 OF 20
Identify the correct statements regarding strong electrolytes and their molar conductivity.
Statements:
1. The intercept of the Λm versus √c plot gives Λ°m.
2. The slope of the plot depends on the temperature and solvent.
3. All electrolytes of a particular type (e.g., 1–1 type like NaCl) have the same value of constant A.
4. Strong electrolytes do not completely dissociate at infinite dilution.
QUESTION 12 OF 20
The plot of Λm versus √c for a strong electrolyte like KCl yields a straight line. What does the negative slope of this line represent?
QUESTION 13 OF 20
For a weak electrolyte at a given concentration c, the degree of dissociation (α) can be approximated by which ratio?
QUESTION 14 OF 20
In the graph of Λm versus √c, the curve for a weak electrolyte like acetic acid compared to a strong electrolyte like KCl:
QUESTION 15 OF 20
The value of limiting molar conductivity (Λ°m) is typically expressed in units of:
QUESTION 16 OF 20
In the expression Λ°m(NaCl) = λ°Na⁺ + λ°Cl⁻, the symbols λ°Na⁺ and λ°Cl⁻ specifically refer to:
QUESTION 17 OF 20
Arrange the following ions in increasing order of their limiting molar conductivity (λ°) in water at 298 K based on standard data trends.
1. OH⁻
2. Na⁺
3. H⁺
4. Ca²⁺
QUESTION 18 OF 20
Match the electrolyte type with the mathematical form of its limiting molar conductivity according to Kohlrausch's law.
| List I | List II |
|---|---|
| 1. NaCl | c. λ°Na⁺ + λ°Cl⁻ |
| 2. MgSO₄ | d. λ°Mg²⁺ + λ°SO₄²⁻ |
| 3. CaCl₂ | a. λ°Ca²⁺ + 2λ°Cl⁻ |
| 4. Al₂(SO₄)₃ | b. 2λ°Al³⁺ + 3λ°SO₄²⁻ |
QUESTION 19 OF 20
Identify the correct equation to calculate Λ°m for acetic acid (HAc) using Kohlrausch's law and strong electrolytes.
QUESTION 20 OF 20
Identify the correct statements regarding determination of the dissociation constant (Ka) of a weak electrolyte.
Statements:
1. Molar conductivity (Λm) at a given concentration must be known.
2. Limiting molar conductivity (Λ°m) must be known.
3. The degree of dissociation α is calculated as Λm/Λ°m.
4. Ka is calculated using .
Test Complete!
Answer Review
1 Identify the relationship used to determine the cell constant (G*) using standard KCl solutions.
�� Cell constant is determined using standard KCl solutions. �� Conductivity of KCl is accurately known. �� Resistance is measured experimentally.
According to NCERT, the cell constant of a conductivity cell is generally determined using standard potassium chloride (KCl) solutions whose conductivity (κ) is accurately known. The conductivity of a solution is related to the cell constant (G*) and resistance (R) by: Rearranging: Thus, after measuring the resistance of the KCl solution and using the known conductivity value, the cell constant can be calculated accurately. Direct measurement of electrode distance and area is difficult due to the irregular nature of platinized platinum electrodes. Therefore, calibration with KCl solution is preferred. The calculated cell constant is then used for conductivity measurements of unknown solutions.
- �� Option A → Gives the inverse relationship.
- �� Option C → Area alone cannot determine cell constant.
- �� Option D → Cell constant is not calculated using conductivity and length alone.
Used – Formula Application
- Application
- Use the relationship κ = G*/R and rearrange.
- Final Logic
- G* = R × κ.
- Cell Constant = Resistance × Conductivity.
2 Why is the measurement of l and A to calculate the cell constant considered inconvenient and unreliable?
�� Platinized electrodes have irregular surfaces. �� Accurate measurement of area is difficult. �� Calibration using KCl is preferred.
The cell constant is theoretically given by: where l is the distance between electrodes and A is their area of cross-section. Although this equation appears simple, NCERT points out that direct determination of l and A is inconvenient and unreliable. The reason is that conductivity cells use platinized platinum electrodes. These electrodes are coated with platinum black, producing a rough and irregular surface. Consequently, the actual effective area of the electrodes cannot be measured accurately. Similarly, determining the exact separation between electrode surfaces is difficult. Because of these practical limitations, the cell constant is determined experimentally using standard KCl solutions of known conductivity rather than by direct geometric measurement.
- �� Option A → Electrode dissolution is not the primary reason.
- �� Option B → AC current passes normally through the conductivity cell.
- �� Option C → Standard KCl solutions are readily available.
Used – NCERT Recall
- Application
- Recall the practical difficulties associated with platinized electrodes.
- Final Logic
- Irregular electrode surfaces make direct measurement unreliable.
- Platinum Black = Difficult Geometry.
3 Identify the correct statements regarding the Wheatstone bridge method for measuring resistance of a conductivity cell.
Statements:
1. The bridge is balanced when no current passes through the detector.
2. Direct current (DC) is used to prevent polarization.
3. An audio-frequency oscillator is used as the AC source.
4. The unknown resistance R₂ is calculated as (R₁R₄)/R₃.
�� Balanced bridge means zero detector current. �� AC is used instead of DC. �� Unknown resistance is calculated from bridge balance.
In a Wheatstone bridge, balance is achieved when no current flows through the detector. Under this condition, the ratio of resistances in one arm equals the ratio in the other arm. Statement 1 is correct because zero detector current indicates balance. Statement 2 is incorrect because direct current causes polarization and electrochemical changes in the solution. Therefore, alternating current is used. Statement 3 is correct because an audio-frequency oscillator supplies the AC current required for accurate measurements. At balance: Rearranging: Hence Statement 4 is also correct. Therefore, Statements 1, 3 and 4 are correct.
- �� Option B → Statement 2 is incorrect.
- �� Option C → Statement 2 is incorrect and Statement 1 is omitted.
- �� Option D → Statement 3 is also correct.
Used – Concept Application
- Application
- Evaluate each statement using Wheatstone bridge principles.
- Final Logic
- Statements 1, 3 and 4 are correct; Statement 2 is false.
- Balance Means No Detector Current.
4 When a Wheatstone bridge is balanced, the measured resistance is in ohms (Ω). The unit of conductance (G), which is the inverse of resistance, is:
�� Conductance is the reciprocal of resistance. �� SI unit is Siemens. �� Ω⁻¹ and S represent the same unit.
Conductance (G) measures the ease with which electric current flows through a conductor. It is defined as the reciprocal of resistance: Since resistance is measured in ohms (Ω), the unit of conductance is: The SI name for Ω⁻¹ is Siemens (S). A larger conductance corresponds to lower resistance and easier current flow. Conductance is widely used in electrochemistry because conductivity calculations are based on conductance measurements. Thus, the SI unit of conductance is Siemens (S), which is equivalent to Ω⁻¹.
- �� Option A → Unit of conductivity.
- �� Option B → Unit of molar conductivity.
- �� Option D → Unit of resistivity.
Used – Formula Recall
- Application
- Use G = 1/R.
- Final Logic
- Inverse of ohm equals Siemens.
- Conductance = Opposite of Resistance.
5 Arrange the following in decreasing order of their typical conductivity at 298.15 K:
1. Copper
2. 0.1 M HCl(aq)
3. Teflon
4. Pure water
�� Metals conduct best. �� Strong acid solutions conduct well. �� Insulators conduct least.
Copper is an excellent metallic conductor with conductivity around S m⁻¹. Therefore, it has the highest conductivity among the given substances. A 0.1 M HCl solution is a strong electrolyte and dissociates almost completely into H⁺ and Cl⁻ ions, giving high ionic conductivity. However, its conductivity is still much lower than that of copper. Pure water contains only a very small concentration of H⁺ and OH⁻ ions produced by self-ionization. Consequently, its conductivity is very low. Teflon is an excellent insulator with extremely low conductivity, making it the poorest conductor among the options. Therefore, the decreasing order is: Copper > 0.1 M HCl(aq) > Pure Water > Teflon Corresponding to: 1 > 2 > 4 > 3
- �� Option B → Places pure water above HCl solution.
- �� Option C → Places HCl above copper.
- �� Option D → Places Teflon above pure water.
Used – Concept Application
- Application
- Compare metallic conductors, electrolytic solutions and insulators.
- Final Logic
- Metal > Strong Electrolyte > Pure Water > Insulator.
- Teflon Conducts Least.
6 Match List-I (Quantity) with List-II (SI Unit).
| List I | List II |
|---|---|
| 1. Resistivity (ρ) | b. Ω m |
| 2. Cell constant (G*) | d. m⁻¹ |
| 3. Conductivity (κ) | a. S m⁻¹ |
| 4. Molar conductivity (Λm) | c. S m² mol⁻¹ A.1-d, 2-a, 3-c, 4-b |
�� Each electrochemical quantity has a distinct SI unit. �� Resistivity and conductivity are reciprocal quantities. �� Molar conductivity includes the mole term.
According to NCERT, resistivity (ρ) is the resistance of a conductor having unit length and unit cross-sectional area. Its SI unit is Ω m. The cell constant (G*) is defined as: and therefore has the SI unit m⁻¹. Conductivity (κ) is the reciprocal of resistivity: Thus, its SI unit is S m⁻¹. Molar conductivity (Λm) is the conductivity associated with one mole of electrolyte and has the SI unit S m² mol⁻¹. Therefore, the correct matching is: 1-b, 2-d, 3-a, 4-c.
- �� Option A → All major units are incorrectly assigned.
- �� Option B → Resistivity and conductivity units are interchanged.
- �� Option C → Cell constant and molar conductivity units are mismatched.
Used – NCERT Recall
- Application
- Recall the SI units of common electrochemical quantities.
- Final Logic
- ρ → Ω m, G* → m⁻¹, κ → S m⁻¹, Λm → S m² mol⁻¹.
- Conductivity Conducts → S m⁻¹.
7 The molar conductivity of a solution increases with a decrease in concentration because:
�� Molar conductivity depends on conductivity and volume. �� Dilution increases the volume containing one mole of electrolyte. �� This effect dominates over the decrease in conductivity.
Molar conductivity is defined as the conductance of the volume of solution containing one mole of electrolyte. It is related to conductivity by: where c is the concentration in mol L⁻¹. When a solution is diluted, conductivity decreases because the number of ions per unit volume decreases. However, the volume containing one mole of electrolyte increases substantially. This increase in volume is much greater than the decrease in conductivity. As a result, molar conductivity increases with dilution. For strong electrolytes, the increase is gradual, while for weak electrolytes it is more pronounced because dilution also increases the degree of dissociation. Therefore, the increase in molar conductivity occurs because the volume effect dominates over the decrease in conductivity.
- �� Option A → Conductivity actually decreases on dilution.
- �� Option C → Number of moles remains unchanged.
- �� Option D → Ionic size does not increase due to dilution.
Used – Concept Application
- Application
- Distinguish between conductivity and molar conductivity.
- Final Logic
- Volume increases faster than conductivity decreases.
- Dilution Lowers κ but Raises Λm.
8 If conductivity (κ) is expressed in S cm⁻¹ and concentration (c) in mol L⁻¹, the formula for molar conductivity (Λm) in S cm² mol⁻¹ is:
�� Molar conductivity depends on conductivity and concentration. �� A factor of 1000 appears because concentration is expressed in mol L⁻¹. �� The formula is commonly used in numerical problems.
When conductivity is expressed in S cm⁻¹ and concentration in mol L⁻¹, NCERT gives the relation: The factor 1000 converts litres into cubic centimetres because: This equation allows determination of molar conductivity directly from conductivity and concentration data. For example, if conductivity decreases upon dilution, the corresponding increase in the volume containing one mole of electrolyte may still cause molar conductivity to increase. This formula is therefore essential for solving electrochemistry numerical problems. Hence, the correct expression is:
- �� Option B → Conductivity is divided by concentration, not multiplied.
- �� Option C → The 1000 factor is incorrectly placed.
- �� Option D → Gives the reciprocal relationship.
Used – Formula Recall
- Application
- Recall the standard molar conductivity formula.
- Final Logic
- Λm = (κ × 1000)/c.
- Thousand on Top, Concentration Below.
9
According to the passage, why does conductivity always decrease with dilution for all electrolytes?
�� Conductivity depends on ions per unit volume. �� Dilution spreads ions over a larger volume. �� Fewer ions per unit volume reduce conductivity.
The passage clearly states that conductivity decreases with dilution for both strong and weak electrolytes. This behavior is explained by the reduction in the number of ions present per unit volume of solution. When water is added, the ions become distributed throughout a larger volume. Although the total number of ions may remain the same, the concentration of ions in any given volume decreases. Since conductivity depends on the availability of charge carriers per unit volume, conductivity decreases. This is different from molar conductivity, which often increases upon dilution. The distinction between conductivity and molar conductivity is an important NCERT concept frequently tested in examinations. Therefore, the correct reason is the decrease in the number of current-carrying ions per unit volume.
- �� Option A → Volume increases rather than decreases.
- �� Option B → Electrode separation remains unchanged.
- �� Option D → Cell constant depends on cell geometry, not concentration.
Used – Passage Analysis
- Application
- Identify the exact explanation given in the passage.
- Final Logic
- Dilution lowers the number of ions per unit volume.
- Dilution = Fewer Ions per Unit Volume.
10
According to the passage, conductivity is equivalent to the conductance of what specific volume of solution?
�� Conductivity is a property of unit dimensions. �� Unit volume is considered. �� Electrodes have unit area and unit separation.
NCERT defines conductivity as the conductance of a solution contained between two electrodes of unit area of cross-section separated by unit distance. Under these conditions, the volume of solution enclosed between the electrodes is one unit volume. This definition allows conductivity to be treated as an intrinsic property of the solution, independent of the actual dimensions of the conductivity cell. Conductivity therefore represents the conductance of a standardized unit volume of solution. The concept is fundamental in electrochemistry because it provides a basis for comparing the conducting abilities of different electrolytic solutions. Hence, conductivity is equivalent to the conductance of one unit volume of solution.
- �� Option A → Conductivity is not defined using one litre specifically.
- �� Option B → One mole relates to molar conductivity.
- �� Option D → One cubic decimeter is not part of the definition.
Used – NCERT Recall
- Application
- Recall the formal NCERT definition of conductivity.
- Final Logic
- Conductivity refers to conductance of a unit volume under standard geometric conditions.
- Unit Area × Unit Length = Unit Volume.
11 Identify the correct statements regarding strong electrolytes and their molar conductivity.
Statements:
1. The intercept of the Λm versus √c plot gives Λ°m.
2. The slope of the plot depends on the temperature and solvent.
3. All electrolytes of a particular type (e.g., 1–1 type like NaCl) have the same value of constant A.
4. Strong electrolytes do not completely dissociate at infinite dilution.
�� Strong electrolytes obey the Debye-Hückel-Onsager equation. �� Λ°m is obtained from the intercept. �� Constant A depends on electrolyte type.
For strong electrolytes, NCERT gives the relation: This equation predicts a straight-line plot between Λm and √c. The intercept at √c = 0 gives the limiting molar conductivity (Λ°m), making Statement 1 correct. The slope of the line is equal to –A. The value of A depends on factors such as the nature of the solvent, temperature and the ionic charges present. Therefore, Statement 2 is also correct. Electrolytes belonging to the same charge type, such as 1–1 electrolytes (NaCl, KCl, etc.), have similar values of A under identical conditions, making Statement 3 correct. Statement 4 is incorrect because strong electrolytes are considered completely dissociated at infinite dilution. Hence, Statements 1, 2 and 3 are correct.
- �� Option B → Statement 4 is incorrect.
- �� Option C → Statement 4 is incorrect.
- �� Option D → Statement 4 is incorrect.
Used – NCERT Recall
- Application
- Recall the Debye-Hückel-Onsager equation and its graphical interpretation.
- Final Logic
- Statements 1, 2 and 3 follow directly from the equation; Statement 4 is false.
- Slope Gives A.
12 The plot of Λm versus √c for a strong electrolyte like KCl yields a straight line. What does the negative slope of this line represent?
�� Strong electrolytes follow a linear equation. �� The slope equals –A. �� A depends on the electrolyte type.
For strong electrolytes, the Debye-Hückel-Onsager equation is: This equation has the form: where: The intercept is Λ°m and the slope is –A. The constant A depends on the type of electrolyte, ionic charges, solvent and temperature. Since the slope of the graph is negative, it is represented by –A. Therefore, the negative slope of the Λm versus √c graph represents the constant A.
- �� Option A → Cell constant is unrelated to the graph slope.
- �� Option B → Λ°m is obtained from the intercept, not slope.
- �� Option D → Degree of dissociation is not represented by the slope.
Used – Formula Application
- Application
- Compare the equation with the straight-line form y = b + mx.
- Final Logic
- Slope = –A and intercept = Λ°m.
- Slope = –A.
13 For a weak electrolyte at a given concentration c, the degree of dissociation (α) can be approximated by which ratio?
�� Degree of dissociation is related to molar conductivity. �� Λm increases with dissociation. �� Infinite dilution corresponds to complete dissociation.
Weak electrolytes are only partially dissociated in solution. At any concentration, the extent of dissociation can be estimated from conductivity measurements. According to NCERT: where: Λm = molar conductivity at concentration c Λ°m = limiting molar conductivity at infinite dilution At infinite dilution, complete dissociation occurs and α approaches 1. Since Λ°m corresponds to complete ionization, the ratio Λm/Λ°m gives the fraction of electrolyte dissociated at a given concentration. This relationship is widely used in calculating dissociation constants of weak electrolytes.
- �� Option A → Inverts the correct expression.
- �� Option C → No such relationship exists.
- �� Option D → Concentration alone cannot determine α.
Used – Formula Recall
- Application
- Recall the conductivity expression for degree of dissociation.
- Final Logic
- α equals observed molar conductivity divided by limiting molar conductivity.
- Actual Over Maximum = α.
14 In the graph of Λm versus √c, the curve for a weak electrolyte like acetic acid compared to a strong electrolyte like KCl:
�� Weak electrolytes dissociate more on dilution. �� Their conductivity increases sharply. �� The graph is curved rather than linear.
Strong electrolytes show nearly complete dissociation at all concentrations. Therefore, their Λm versus √c plots are approximately straight lines. Weak electrolytes behave differently. As dilution increases, the degree of dissociation rises significantly. This produces a rapid increase in the number of ions available for conduction. Consequently, the molar conductivity increases steeply at lower concentrations. The resulting graph is curved and non-linear rather than a straight line. Because of this non-linearity, Λ°m cannot be obtained by simple extrapolation of the graph to zero concentration. Therefore, the characteristic feature of weak electrolytes is a steep upward curve at low concentrations.
- �� Option A → Weak electrolyte plots are not linear.
- �� Option B → Direct interception is not possible.
- �� Option C → Λm increases rather than decreases.
Used – Concept Application
- Application
- Compare conductivity behavior of strong and weak electrolytes.
- Final Logic
- Increasing dissociation produces a steep non-linear curve.
- Weak Electrolyte = Curved Rise.
15 The value of limiting molar conductivity (Λ°m) is typically expressed in units of:
�� Limiting molar conductivity is a form of molar conductivity. �� It includes area and mole terms. �� The commonly used unit is S cm² mol⁻¹.
Limiting molar conductivity (Λ°m) is the molar conductivity of an electrolyte at infinite dilution. Since it is a molar conductivity quantity, it has the same unit as molar conductivity. When conductivity is expressed in S cm⁻¹ and concentration in mol L⁻¹, the commonly used unit becomes: This unit indicates the conductance contribution associated with one mole of electrolyte. Limiting molar conductivity is especially important in Kohlrausch's law and in determining dissociation constants of weak electrolytes. Therefore, the standard unit of Λ°m is S cm² mol⁻¹.
- �� Option A → Unit of conductivity.
- �� Option B → Also a conductivity unit.
- �� Option D → Unit related to resistivity.
Used – NCERT Recall
- Application
- Recall the standard unit of molar conductivity.
- Final Logic
- Λ°m uses the same unit as molar conductivity.
- Molar Conductivity = cm² per Mole.
16 In the expression Λ°m(NaCl) = λ°Na⁺ + λ°Cl⁻, the symbols λ°Na⁺ and λ°Cl⁻ specifically refer to:
�� Kohlrausch's law uses ionic contributions. �� Each ion contributes independently. �� λ° values represent limiting ionic conductivities.
According to Kohlrausch's law of independent migration of ions, each ion contributes a definite value to the limiting molar conductivity of an electrolyte at infinite dilution. In the expression: the symbols λ°Na⁺ and λ°Cl⁻ represent the limiting molar conductivities of sodium ions and chloride ions respectively. At infinite dilution, ions move independently without significant interionic interactions. Therefore, the total limiting molar conductivity of an electrolyte is obtained by adding the individual ionic contributions. These ionic conductivities are experimentally determined quantities and form the basis of Kohlrausch's law. Hence, λ°Na⁺ and λ°Cl⁻ specifically denote limiting molar conductivities of the individual ions.
- �� Option A → λ° does not represent concentration.
- �� Option C → Equivalent weight is unrelated to λ°.
- �� Option D → Ionic mobility influences conductivity but λ° itself represents limiting ionic conductivity.
Used – NCERT Recall
- Application
- Recall the symbols used in Kohlrausch's law.
- Final Logic
- λ° values are limiting ionic conductivities.
- λ° = Ionic Contribution at Infinite Dilution.
17 Arrange the following ions in increasing order of their limiting molar conductivity (λ°) in water at 298 K based on standard data trends.
1. OH⁻
2. Na⁺
3. H⁺
4. Ca²⁺
�� H⁺ and OH⁻ possess exceptionally high conductivities. �� Metal ions have comparatively lower values. �� Proton transport is exceptionally fast.
The limiting molar conductivity of ions depends on their mobility in aqueous solution. Hydrogen ions and hydroxide ions exhibit exceptionally high conductivities because they move through the solution via proton-transfer mechanisms rather than simple ionic migration. Given ions: 1. OH⁻ 2. Na⁺ 3. H⁺ 4. Ca²⁺ Typical conductivity trend: Hydrogen ion has the highest limiting molar conductivity among common ions, while hydroxide ion is also exceptionally high. Sodium ion possesses lower mobility because of extensive hydration. Calcium ion, although doubly charged, has conductivity greater than sodium ion under standard conditions. Therefore, the increasing order is: Na⁺ < Ca²⁺ < OH⁻ < H⁺ Corresponding to: 2 < 4 < 1 < 3
- �� Option B → Gives the reverse trend for highly mobile ions.
- �� Option C → Places H⁺ below OH⁻ incorrectly.
- �� Option D → Places OH⁻ and H⁺ before metal ions.
Used – NCERT Recall
- Application
- Recall standard ionic conductivity trends.
- Final Logic
- Na⁺ < Ca²⁺ < OH⁻ < H⁺.
- Metal Ions Lower.
18 Match the electrolyte type with the mathematical form of its limiting molar conductivity according to Kohlrausch's law.
| List I | List II |
|---|---|
| 1. NaCl | c. λ°Na⁺ + λ°Cl⁻ |
| 2. MgSO₄ | d. λ°Mg²⁺ + λ°SO₄²⁻ |
| 3. CaCl₂ | a. λ°Ca²⁺ + 2λ°Cl⁻ |
| 4. Al₂(SO₄)₃ | b. 2λ°Al³⁺ + 3λ°SO₄²⁻ |
�� Ionic contributions are added. �� Stoichiometric coefficients are included. �� Kohlrausch's law applies at infinite dilution.
According to Kohlrausch's law: For NaCl: For MgSO₄: For CaCl₂: For Al₂(SO₄)₃: Thus the correct matching is: 1-c, 2-d, 3-a, 4-b.
- �� Option B → Incorrect stoichiometric assignments.
- �� Option C → MgSO₄ and CaCl₂ are mismatched.
- �� Option D → Multiple electrolyte formulas are incorrectly assigned.
Used – Formula Application
- Application
- Apply Kohlrausch's law using ionic stoichiometry.
- Final Logic
- Multiply ionic conductivities by their stoichiometric coefficients.
- Add Ionic Contributions with Their Coefficients.
19 Identify the correct equation to calculate Λ°m for acetic acid (HAc) using Kohlrausch's law and strong electrolytes.
�� Kohlrausch's law allows indirect determination. �� Ionic contributions are added and subtracted. �� Useful for weak electrolytes.
The limiting molar conductivity of acetic acid cannot be obtained directly by extrapolation because acetic acid is a weak electrolyte. Therefore, Kohlrausch's law is used. Using ionic contributions: Adding the first two equations and subtracting the third eliminates Na⁺ and Cl⁻: Hence Option D is correct.
- �� Option A → Does not yield HAc conductivity correctly.
- �� Option B → Removes H⁺ contribution incorrectly.
- �� Option C → Gives an incorrect ionic cancellation.
Used – Substitution
- Application
- Add and subtract ionic conductivity equations.
- Final Logic
- Cancel common ions and retain H⁺ and Ac⁻.
- HCl + NaAc − NaCl = HAc.
20 Identify the correct statements regarding determination of the dissociation constant (Ka) of a weak electrolyte.
Statements:
1. Molar conductivity (Λm) at a given concentration must be known.
2. Limiting molar conductivity (Λ°m) must be known.
3. The degree of dissociation α is calculated as Λm/Λ°m.
4. Ka is calculated using .
�� Conductivity measurements give α. �� Λ°m is needed for comparison. �� Ka is calculated using Ostwald's dilution law.
The dissociation constant of a weak electrolyte can be determined using conductivity measurements. First, the molar conductivity at a given concentration (Λm) is measured experimentally. The limiting molar conductivity (Λ°m) is then obtained using Kohlrausch's law. The degree of dissociation is calculated as: After determining α, Ostwald's dilution law is applied: where c is the concentration of the electrolyte. Thus, determination of Ka requires knowledge of Λm, Λ°m, α and the above equation. Therefore, all four statements are correct.
- �� Option A → Statements 3 and 4 are also correct.
- �� Option B → Statements 1 and 2 are also required.
- �� Option D → Statement 2 is also essential.
Used – Concept Application
- Application
- Follow the complete sequence used for Ka determination.
- Final Logic
- Λm → Λ°m → α → Ka.
- Obtain Ka.
