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At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 x 102 S cm2 mol-1. At 298 K, for an aqueous solution of the acid, the degree of dissociation is and the molar conductivity is y x 102 S cm2 mol-1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y x 102 S cm2 mol-1.
Q. The value of y is _______.
Correct answer is '0.86'. Can you explain this answer?
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At 298 K, the limiting molar conductivity of a weak monobasic acid is ...
The molar conductivity of an electrolyte solution is a measure of its ability to conduct electricity. It depends on the concentration and degree of dissociation of the electrolyte.
Given:
Limiting molar conductivity of the weak monobasic acid at 298 K = 4 x 10^2 S cm^2 mol^-1
Molar conductivity of the acid solution at 298 K = y x 10^2 S cm^2 mol^-1
Molar conductivity of the diluted acid solution at 298 K = 3y x 10^2 S cm^2 mol^-1
To find the value of y, we need to understand the relationship between molar conductivity, degree of dissociation, and dilution.
1. The degree of dissociation (α) is the fraction of the acid molecules that dissociate into ions in the solution. It can be calculated using the formula:
α = (molar conductivity of the acid solution / limiting molar conductivity of the acid) * 100
α = (y x 10^2 / 4 x 10^2) * 100
α = (y/4) * 100
2. Upon dilution, the molar conductivity of the solution increases. This is because dilution reduces the concentration of the electrolyte, allowing for more free movement of ions. The relationship between molar conductivity and dilution is given by:
Molar conductivity after dilution / Molar conductivity before dilution = Dilution factor
(3y x 10^2) / (y x 10^2) = 20
3y / y = 20
3 = 20
y = 20/3
y ≈ 0.86
Therefore, the value of y is approximately 0.86.
Given:
Limiting molar conductivity of the weak monobasic acid at 298 K = 4 x 10^2 S cm^2 mol^-1
Molar conductivity of the acid solution at 298 K = y x 10^2 S cm^2 mol^-1
Molar conductivity of the diluted acid solution at 298 K = 3y x 10^2 S cm^2 mol^-1
To find the value of y, we need to understand the relationship between molar conductivity, degree of dissociation, and dilution.
1. The degree of dissociation (α) is the fraction of the acid molecules that dissociate into ions in the solution. It can be calculated using the formula:
α = (molar conductivity of the acid solution / limiting molar conductivity of the acid) * 100
α = (y x 10^2 / 4 x 10^2) * 100
α = (y/4) * 100
2. Upon dilution, the molar conductivity of the solution increases. This is because dilution reduces the concentration of the electrolyte, allowing for more free movement of ions. The relationship between molar conductivity and dilution is given by:
Molar conductivity after dilution / Molar conductivity before dilution = Dilution factor
(3y x 10^2) / (y x 10^2) = 20
3y / y = 20
3 = 20
y = 20/3
y ≈ 0.86
Therefore, the value of y is approximately 0.86.
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At 298 K, the limiting molar conductivity of a weak monobasic acid is ...
When solution is diuted 20 times with water α2 = 3α1
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At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 x 102 S cm2 mol-1. At 298 K, for an aqueous solution of the acid, the degree of dissociation is and the molar conductivity is y x 102 S cm2 mol-1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y x 102 S cm2 mol-1.Q. The value of y is _______.Correct answer is '0.86'. Can you explain this answer?
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At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 x 102 S cm2 mol-1. At 298 K, for an aqueous solution of the acid, the degree of dissociation is and the molar conductivity is y x 102 S cm2 mol-1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y x 102 S cm2 mol-1.Q. The value of y is _______.Correct answer is '0.86'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 x 102 S cm2 mol-1. At 298 K, for an aqueous solution of the acid, the degree of dissociation is and the molar conductivity is y x 102 S cm2 mol-1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y x 102 S cm2 mol-1.Q. The value of y is _______.Correct answer is '0.86'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 x 102 S cm2 mol-1. At 298 K, for an aqueous solution of the acid, the degree of dissociation is and the molar conductivity is y x 102 S cm2 mol-1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y x 102 S cm2 mol-1.Q. The value of y is _______.Correct answer is '0.86'. Can you explain this answer?.
At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 x 102 S cm2 mol-1. At 298 K, for an aqueous solution of the acid, the degree of dissociation is and the molar conductivity is y x 102 S cm2 mol-1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y x 102 S cm2 mol-1.Q. The value of y is _______.Correct answer is '0.86'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 x 102 S cm2 mol-1. At 298 K, for an aqueous solution of the acid, the degree of dissociation is and the molar conductivity is y x 102 S cm2 mol-1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y x 102 S cm2 mol-1.Q. The value of y is _______.Correct answer is '0.86'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 x 102 S cm2 mol-1. At 298 K, for an aqueous solution of the acid, the degree of dissociation is and the molar conductivity is y x 102 S cm2 mol-1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y x 102 S cm2 mol-1.Q. The value of y is _______.Correct answer is '0.86'. Can you explain this answer?.
Solutions for At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 x 102 S cm2 mol-1. At 298 K, for an aqueous solution of the acid, the degree of dissociation is and the molar conductivity is y x 102 S cm2 mol-1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y x 102 S cm2 mol-1.Q. The value of y is _______.Correct answer is '0.86'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 x 102 S cm2 mol-1. At 298 K, for an aqueous solution of the acid, the degree of dissociation is and the molar conductivity is y x 102 S cm2 mol-1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y x 102 S cm2 mol-1.Q. The value of y is _______.Correct answer is '0.86'. Can you explain this answer?, a detailed solution for At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 x 102 S cm2 mol-1. At 298 K, for an aqueous solution of the acid, the degree of dissociation is and the molar conductivity is y x 102 S cm2 mol-1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y x 102 S cm2 mol-1.Q. The value of y is _______.Correct answer is '0.86'. Can you explain this answer? has been provided alongside types of At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 x 102 S cm2 mol-1. At 298 K, for an aqueous solution of the acid, the degree of dissociation is and the molar conductivity is y x 102 S cm2 mol-1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y x 102 S cm2 mol-1.Q. The value of y is _______.Correct answer is '0.86'. Can you explain this answer? theory, EduRev gives you an
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