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Kohlrausch Law Questions

Kohlrausch Law directs the electrolyte’s limiting molar conductivity with its constituent ions. It displays that at infinite dilution equivalent conductivity of an electrolyte is equivalent to the sum of the conductances of the cations and anions. For example, the equivalent conductivity of CH3COONa at equivalent dilution would be equivalent to the sum of the conductances of the acetate and sodium ion.

𝛬mº CH3COONa = 𝛬mº CH3COO + 𝛬mº Na +

Definition: Kohlrausch’s law states that the equivalent conductivity of an electrolyte at infinite dilution is equivalent to the sum of the conductances of cations and anions.

Kohlrausch Law Chemistry Questions with Solutions

Q1. Equivalent conductance of a strong electrolyte increases on dilution due to

(a) An increase in the number of ions and the ionic mobility of solution

(b) Complete dilution of the electrolyte at standard dilution

(c ) An increase in the ionic mobility of solution

(d) None of the above

Answer: (c ) Equivalent conductance of a strong electrolyte increases on dilution due to an increase in the ionic mobility of solution.

Q2. The molar conductivity of an ionic solution depends on

(a) Concentration of electrolytes in solution

(b) Distance between electrodes

(c ) Surface area of electrodes

(d) None of the above

Answer: (a) The molar conductivity of an ionic solution depends on the concentration of electrolytes in solution.

Q3. The molar conductance of a solution ___________ with dilution while its specific conductance ________ with dilution.

(a) Decreases, Increases

(b) Increases, Decreases

(c ) Decreases, Decreases

(d) Increases, Increases

Answer: (a) The molar conductance of a solution increases with dilution while its specific conductance decreases with dilution.

Q4. Kohlrausch’s law states that at

(a) Infinite dilution, the equivalent conductivity of an electrolyte is equivalent to the sum of the conductances of the cations and anions

(b) Finite dilution, the equivalent conductivity of an electrolyte is equivalent to the sum of the conductances of the cations and anions.

(c ) Both (a) and (b)

(d) None of the above

Answer: (a) Kohlrausch’s law states that at infinite dilution, the equivalent conductivity of an electrolyte is equivalent to the sum of the conductances of the cations and the anions

Q5. Which of the following statements is correct for an electrolytic solution upon dilution?

(a) Conductivity increase on dilution

(b) Conductivity decrease on dilution

(c ) Molar conductance decreases, but equivalent conductance increases on dilution

(d) Molar conductance increases, but equivalent conductance decreases on dilution

Answer: (b) When an electrolytic solution is diluted, it’s concentration decreases. Due to this, the conductance increases, and the conductivity decreases.

Q6. What is Kohlrausch’s law?

Answer: Kohlrausch’s law states that at infinite dilution equivalent conductivity of an electrolyte is equivalent to the sum of the conductances of the cations and anions.

Q7. What are the applications of Kohlrausch’s law?

Answer: Kohlrausch’s law states that at infinite dilution equivalent conductivity of an electrolyte is equivalent to the sum of the conductances of the cations and anions. There are a lot of applications of Kohlrausch’s law.

A few of them are mentioned below.

1. It can be used to calculate the molar conductance at infinite dilution for a weak electrolyte.

2. It can be used to calculate the degree of dissociation at infinite dilution for a weak electrolyte.

3. It can be used to calculate the dissociation constant at infinite dilution for a weak electrolyte.

4. It can be used to calculate the solubility of the sparingly soluble salt.

Q8. The 𝛬mº for sodium iodide, sodium acetate, and magnesium acetate solution are 12.69, 9.10 and 18.78 S cm2 mol-1, respectively, at 298 K. Calculate 𝛬mº for magnesium iodide.

Answer:

We know that

𝛬mº (NaI) = 12.69 S cm2 mol-1

𝛬mº (CH3COONa) = 9.10 S cm2 mol-1

𝛬mº (MgI2) = 18.78 S cm2 mol-1

According to Kohlrausch’s law

𝛬mº (MgI2) = 𝛬mº [CH3(COO)]2Mg + 2 𝛬mº (NaI) – 2 𝛬mº (CH3COONa)

𝛬mº (MgI2) = (18.78 + 2 X 12.69 – 2 X 9.10) S cm2 mol-1

𝛬mº (MgI2) = (18.78 + 25.38 – 18.20) S cm2 mol-1

𝛬mº (MgI2) = (44.16 – 18.20) S cm2 mol-1

𝛬mº (MgI2) = 25.96 S cm2 mol-1

Q9. Calculate 𝛬mº for CaCl2 and MgSO4 from the following data:

𝛬mº (Ca2+) = 119.0 S cm2 mol-1, 𝛬mº (Mg2+) = 106.0 S cm2 mol-1, 𝛬mº (Cl) = 76.3 S cm2 mol-1 and 𝛬mº (SO42- ) = 160.05 S cm2 mol-1

Answer:

We know that,

𝛬mº (Ca2+) = 119.0 S cm2 mol-1

𝛬mº (Mg2+) = 106.0 S cm2 mol-1

𝛬mº (Cl) = 76.3 S cm2 mol-1

𝛬mº (SO42- ) = 160.05 S cm2 mol-1

According to Kohlrausch’s law

𝛬mº (CaCl2) = 𝛬mº (Ca2+) + 2 X [𝛬mº (Cl)]

𝛬mº (CaCl2) = [119.0 + 2 X 76.3] S cm2 mol-1

𝛬mº (CaCl2) = [119.0 + 152.6] S cm2 mol-1

𝛬mº (CaCl2) = 271.6 S cm2 mol-1

And

𝛬mº (MgSO4) = 𝛬mº (Mg2+) + 𝛬mº (SO42- )

𝛬mº (MgSO4) = [106.0 + 160.05] S cm2 mol-1

𝛬mº (MgSO4) = 266.05 S cm2 mol-1

Q10. The 𝛬mº for sodium acetate, HCl, and NaCl are 91.0, 425.9 and 126.4 S cm2 mol-1, respectively, at 298 K. Calculate 𝛬mº for CH3COOH.

Answer:

We know that,

𝛬mº (CH3COONa) = 91.0 S cm2 mol-1

𝛬mº (HCl) = 425.9 S cm2 mol-1

𝛬mº (NaCl) = 126.4 S cm2 mol-1

According to Kohlrausch’s law

𝛬mº CH3COOH = 𝛬mº CH3COONa + 𝛬mº HCl – 𝛬mº NaCl

𝛬mº CH3COOH = (91.0 + 425.9 – 126.4) S cm2 mol-1

𝛬mº CH3COOH = 390.5 S cm2 mol-1

Q11. The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm2 mol −1. Calculate the conductivity of this solution.

Answer: Molar conductivity (𝜆m) = 138.9 S cm2 mol −1

Concentration = 1.5 M

Molar conductivity = Conductivity / Concentration

Conductivity = Molar conductivity X Concentration

Conductivity = (138.9 S cm2 mol −1 X 1.5 M) / 1000 cm3 L-1

Conductivity = 0.208 S cm -1

Q12. The resistance and conductivity of a cell containing 0.001 M KCl solution at 298 K are 1500 Ω and 1.46 X 10−4 S cm−1, respectively. What is the cell constant of the cell?

Answer:

Resistance = 1500 Ω

Conductivity = 1.46 X 10−4 S cm−1

Concentration = 0.001 M

Temperature = 298 K

We know that,

Conductivity = Cell constant / Resistance

1.46 X 10−4 S cm−1 = Cell constant / 1500 Ω

Cell constant = 1.46 X 10−4 S cm−1 X 1500 Ω

Cell constant = 0.219 cm -1

Q13. The conductivity of the 0.20 M solution of KCl at 298 K is 0.0248 S cm−1. Calculate its molar conductivity.

Answer:

Conductivity = 0.0248 S cm−1

Concentration = 0.20 M

Temperature = 298 K

Molar conductivity (𝜆m) = (Conductivity X 1000) / C

Molar conductivity (𝜆m) = (0.0248 S cm−1 X 1000) / 0.20 mol cm -3

Molar conductivity (𝜆m) = 24.8 S cm−1 / 0.20 mol cm -3

Molar conductivity (𝜆m) = 124 S cm2 mol−1

Q14. The molar conductivity of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91 S cm2 mol −1, respectively. What will be the molar conductivity of CH3COOH at infinite dilution?

Answer:

We know that,

𝛬mº (NaCl) = 126.45 S cm2 mol −1

𝛬mº (HCl) = 426.16 S cm2 mol −1

𝛬mº (CH3COONa) = 91 S cm2 mol −1

According to Kohlrausch’s law

𝛬mº CH3COOH = 𝛬mº CH3COONa + 𝛬mº HCl – 𝛬mº NaCl

𝛬mº CH3COOH = (91 + 426.16 – 126.45) S cm2 mol −1

𝛬mº CH3COOH = (91 + 426.16 – 126.45) S cm2 mol −1

𝛬mº CH3COOH = (517.16 – 126.45) S cm2 mol −1

𝛬mº CH3COOH = 390.71 S cm2 mol −1

Q15. Match the following.

Column A

Column B

Kohlrausch’s law

K α = Cα2 / 1 – α

Molar conductivity

α = Λ m / Λ m o

Degree of dissociation

Λ m = k / C

Dissociation constant

Λ eq o = Λ c o + Λ a o

Answer:

Column A

Column B

Kohlrausch’s law

Λ eq o = Λ c o + Λ a o

Molar conductivity

Λ m = k / C

Degree of dissociation

α = Λ m / Λ m o

Dissociation constant

K α = Cα2 / 1 – α

Practise Questions on Kohlrausch Law

Q1. Why can 𝛬mº for CH3COOH not be determined experimentally?

Q2. Why does the conductivity of a solution decrease with dilution?

Q3. The conductivity of 0.00241 M acetic acid is 7.896 X 10-5 S cm-1. Calculate its molar conductivity, and if 𝛬mº for acetic acid is 390.5 S cm2 mol -1, what is its dissociation constant?

Q4. Three electrolytic cells A, B, and C, containing a solution of ZnSO4, AgNO3 and CuSO4, respectively all connected in series. A steady current of 1.5 amperes was passed through then until 1.45 g of silver was deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Q5. Suggest a way to determine the 𝛬mº of CH3COOH.

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