chemistry : class12 Dilution Effect on Molar Conductivity : Conductance in electrolytic solutions : Electrochemistry MCQ Questions and Answers
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Dilution Effect on Molar Conductivity
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MCQ Questions
1. A strong electrolyte solution has conductivity 2.00 × 10^-3 S cm^-1 at 0.020 mol L^-1. On dilution to 0.0020 mol L^-1, its conductivity becomes 5.00 × 10^-4 S cm^-1. What is the ratio of its molar conductivity after dilution to that before dilution?
A.
1.25
B.
2.50
C.
4.00
D.
0.25
Correct Answer:
2.50
Explanation:
Molar conductivity is Λm = 1000κ/C. Therefore, Λm,2/Λm,1 = (κ2/κ1)(C1/C2) = (5.00 × 10^-4 / 2.00 × 10^-3)(0.020/0.0020) = 0.25 × 10 = 2.50. Although conductivity decreases on dilution, molar conductivity increases because concentration decreases more strongly.
2. For a strong electrolyte, the dependence of molar conductivity on concentration is given by Λm = Λm° − K√C. Its molar conductivity is 120 S cm^2 mol^-1 at 0.010 mol L^-1 and 100 S cm^2 mol^-1 at 0.040 mol L^-1. What is its limiting molar conductivity?
A.
120 S cm^2 mol^-1
B.
130 S cm^2 mol^-1
C.
140 S cm^2 mol^-1
D.
160 S cm^2 mol^-1
Correct Answer:
140 S cm^2 mol^-1
Explanation:
At 0.010 M, √C = 0.10, and at 0.040 M, √C = 0.20. The change in molar conductivity is 20 = K(0.20 − 0.10), so K = 200. Hence Λm° = 120 + 200(0.10) = 140 S cm^2 mol^-1. This relation describes the increase in molar conductivity on dilution of a strong electrolyte.
3. A weak electrolyte has limiting molar conductivity 400 S cm^2 mol^-1. Its molar conductivity changes from 20 S cm^2 mol^-1 at 0.100 mol L^-1 to 100 S cm^2 mol^-1 at 0.010 mol L^-1. By what factor does its degree of ionization increase on this dilution?
A.
2
B.
4
C.
5
D.
20
Correct Answer:
5
Explanation:
For a weak electrolyte, the degree of ionization is α = Λm/Λm°. Initially, α1 = 20/400 = 0.05, while after dilution α2 = 100/400 = 0.25. Thus, α2/α1 = 0.25/0.05 = 5. Dilution increases ionization and therefore increases molar conductivity.
4. The molar conductivity of a strong electrolyte is 150 S cm^2 mol^-1 at 0.010 mol L^-1 and 130 S cm^2 mol^-1 at 0.040 mol L^-1. Assuming the relation Λm = Λm° − K√C, what is its molar conductivity at 0.0025 mol L^-1?
A.
155 S cm^2 mol^-1
B.
160 S cm^2 mol^-1
C.
165 S cm^2 mol^-1
D.
170 S cm^2 mol^-1
Correct Answer:
160 S cm^2 mol^-1
Explanation:
The difference in Λm is 20 for a difference in √C of 0.20 − 0.10 = 0.10, giving K = 200. Thus Λm° = 150 + 200(0.10) = 170 S cm^2 mol^-1. At 0.0025 M, √C = 0.05, so Λm = 170 − 200(0.05) = 160 S cm^2 mol^-1.
5. For two solutions of the same weak electrolyte, the molar conductivities are 40 S cm^2 mol^-1 at 0.050 mol L^-1 and 160 S cm^2 mol^-1 at 0.0050 mol L^-1. If the limiting molar conductivity is 200 S cm^2 mol^-1, what is the ratio of their degrees of ionization, α at 0.0050 M divided by α at 0.050 M?
A.
2
B.
4
C.
5
D.
8
Correct Answer:
4
Explanation:
Since α = Λm/Λm°, α at 0.050 M is 40/200 = 0.20, and α at 0.0050 M is 160/200 = 0.80. Their ratio is 0.80/0.20 = 4. Dilution substantially increases ionization in a weak electrolyte.
6. A weak electrolyte has Λm° = 250 S cm^2 mol^-1. At a certain concentration its degree of ionization is 0.20. If dilution increases its degree of ionization to 0.50, what is the percentage increase in molar conductivity?
A.
30%
B.
60%
C.
150%
D.
250%
Correct Answer:
150%
Explanation:
Because Λm = αΛm°, the initial molar conductivity is 0.20 × 250 = 50 S cm^2 mol^-1, and the final value is 0.50 × 250 = 125 S cm^2 mol^-1. Percentage increase = [(125 − 50)/50] × 100 = 150%. The increase results from enhanced ionization on dilution.
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