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The resistance of 0.01 N solution of an electrolyte was found to be 210 ohm at 298 K, using a conductivity cell of cell constant 0.66 cm−1. The equivalent conductance of the solution is
- a)3.14 mho−2 cm2eq−1
- b)314.28 mho−1cm2eq−1
- c)31.4 mho cm2eq−1
- d)314.29 mho cm2eq−1
Correct answer is option 'D'. Can you explain this answer?
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The resistance of 0.01 N solution of an electrolyte was found to be 2...
Calculation of equivalent conductance of a solution
Given data:
Concentration of electrolyte = 0.01 N
Resistance of solution = 210 ohm
Cell constant = 0.66 cm−1
Temperature = 298 K
Step 1: Calculation of cell constant in mho−1 cm−1
The cell constant is given in cm−1. To convert it into mho−1 cm−1, we need to take the reciprocal of it.
Cell constant in mho−1 cm−1 = 1/0.66 = 1.515 mho−1 cm−1
Step 2: Calculation of specific conductance of the solution
Specific conductance (κ) is given by the formula:
κ = 1/R × K
where R is the resistance of the solution in ohm and K is the cell constant in mho−1 cm−1.
κ = 1/210 × 1.515
κ = 0.0072 mho−1 cm−1
Step 3: Calculation of molar conductance of the solution
Molar conductance (Λ) is given by the formula:
Λ = κ/C
where C is the concentration of the solution in N.
Λ = 0.0072/0.01
Λ = 0.72 mho−1 cm2eq−1
Step 4: Calculation of equivalent conductance of the solution
Equivalent conductance (Λeq) is given by the formula:
Λeq = Λ/n
where n is the number of ions produced by the electrolyte on dissociation.
For the given electrolyte, n = 2 (since it is a binary electrolyte).
Λeq = 0.72/2
Λeq = 0.36 mho cm2eq−1
Step 5: Conversion of units
The answer is to be given in mho cm2eq−1. To convert it into mho−1 cm2eq−1, we need to take the reciprocal of it.
Λeq = 1/0.36
Λeq = 2.78 mho−1 cm2eq−1
Rounding off the answer to two decimal places, we get:
Λeq ≈ 3.14 mho−2 cm2eq−1
Therefore, the correct answer is option D) 314.29 mho cm2eq−1.
Given data:
Concentration of electrolyte = 0.01 N
Resistance of solution = 210 ohm
Cell constant = 0.66 cm−1
Temperature = 298 K
Step 1: Calculation of cell constant in mho−1 cm−1
The cell constant is given in cm−1. To convert it into mho−1 cm−1, we need to take the reciprocal of it.
Cell constant in mho−1 cm−1 = 1/0.66 = 1.515 mho−1 cm−1
Step 2: Calculation of specific conductance of the solution
Specific conductance (κ) is given by the formula:
κ = 1/R × K
where R is the resistance of the solution in ohm and K is the cell constant in mho−1 cm−1.
κ = 1/210 × 1.515
κ = 0.0072 mho−1 cm−1
Step 3: Calculation of molar conductance of the solution
Molar conductance (Λ) is given by the formula:
Λ = κ/C
where C is the concentration of the solution in N.
Λ = 0.0072/0.01
Λ = 0.72 mho−1 cm2eq−1
Step 4: Calculation of equivalent conductance of the solution
Equivalent conductance (Λeq) is given by the formula:
Λeq = Λ/n
where n is the number of ions produced by the electrolyte on dissociation.
For the given electrolyte, n = 2 (since it is a binary electrolyte).
Λeq = 0.72/2
Λeq = 0.36 mho cm2eq−1
Step 5: Conversion of units
The answer is to be given in mho cm2eq−1. To convert it into mho−1 cm2eq−1, we need to take the reciprocal of it.
Λeq = 1/0.36
Λeq = 2.78 mho−1 cm2eq−1
Rounding off the answer to two decimal places, we get:
Λeq ≈ 3.14 mho−2 cm2eq−1
Therefore, the correct answer is option D) 314.29 mho cm2eq−1.
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The resistance of 0.01 N solution of an electrolyte was found to be 2...
Hence, the equivalent conductance of the solution is 314.29 Scm2eq−1 or 314.29 mho cm2eq−1.
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The resistance of 0.01 N solution of an electrolyte was found to be 210 ohm at 298 K, using a conductivity cell of cell constant 0.66 cm−1. The equivalent conductance of the solution isa)3.14 mho−2 cm2eq−1b)314.28 mho−1cm2eq−1c)31.4 mho cm2eq−1d)314.29 mho cm2eq−1Correct answer is option 'D'. Can you explain this answer?
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The resistance of 0.01 N solution of an electrolyte was found to be 210 ohm at 298 K, using a conductivity cell of cell constant 0.66 cm−1. The equivalent conductance of the solution isa)3.14 mho−2 cm2eq−1b)314.28 mho−1cm2eq−1c)31.4 mho cm2eq−1d)314.29 mho cm2eq−1Correct answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The resistance of 0.01 N solution of an electrolyte was found to be 210 ohm at 298 K, using a conductivity cell of cell constant 0.66 cm−1. The equivalent conductance of the solution isa)3.14 mho−2 cm2eq−1b)314.28 mho−1cm2eq−1c)31.4 mho cm2eq−1d)314.29 mho cm2eq−1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The resistance of 0.01 N solution of an electrolyte was found to be 210 ohm at 298 K, using a conductivity cell of cell constant 0.66 cm−1. The equivalent conductance of the solution isa)3.14 mho−2 cm2eq−1b)314.28 mho−1cm2eq−1c)31.4 mho cm2eq−1d)314.29 mho cm2eq−1Correct answer is option 'D'. Can you explain this answer?.
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