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The molar conductivity of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91 S cm2 mol–1 respectively. The molar conductivity of CH3COOH at infinite dilution is:
- a)201.28 S cm2 mol–1
- b)698.28 S cm2 mol–1
- c)390.71 S cm2 mol–1
- d)540.48 S cm2 mol–1
Correct answer is option 'C'. Can you explain this answer?
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The molar conductivity of NaCl, HCl and CH3COONa at infinite dilution ...
To find the molar conductivity of CH3COOH at infinite dilution, we can make use of the Kohlrausch's law of independent migration of ions. According to this law, the molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of its constituent ions.
Given:
Molar conductivity of NaCl at infinite dilution = 126.45 S cm2 mol1
Molar conductivity of HCl at infinite dilution = 426.16 S cm2 mol1
Molar conductivity of CH3COONa at infinite dilution = 91 S cm2 mol1
We need to find the molar conductivity of CH3COOH.
Step 1: Write the dissociation equation for CH3COOH.
CH3COOH ⇌ CH3COO- + H+
Step 2: Determine the number of ions produced by the dissociation of CH3COOH.
In the dissociation equation, one CH3COOH molecule produces one CH3COO- ion and one H+ ion. So, the total number of ions produced is 2.
Step 3: Apply Kohlrausch's law to calculate the molar conductivity of CH3COOH.
According to Kohlrausch's law, the molar conductivity of CH3COOH can be calculated using the equation:
Λ(CH3COOH) = Λ(CH3COO-) + Λ(H+)
Given that the molar conductivity of NaCl is 126.45 S cm2 mol1 and it dissociates into Na+ and Cl- ions, we can write:
Λ(NaCl) = Λ(Na+) + Λ(Cl-)
Similarly, the molar conductivity of HCl is 426.16 S cm2 mol1 and it dissociates into H+ and Cl- ions, so:
Λ(HCl) = Λ(H+) + Λ(Cl-)
Using the above equations and the given molar conductivities, we can rearrange the equation to solve for the molar conductivity of CH3COOH:
Λ(CH3COOH) = Λ(NaCl) - Λ(Cl-) + Λ(HCl) - Λ(H+)
Λ(CH3COOH) = 126.45 - 91 + 426.16 - 426.16
Simplifying the equation gives:
Λ(CH3COOH) = 201.28 S cm2 mol1
Therefore, the molar conductivity of CH3COOH at infinite dilution is 201.28 S cm2 mol1.
Hence, option 'a' is the correct answer.
Given:
Molar conductivity of NaCl at infinite dilution = 126.45 S cm2 mol1
Molar conductivity of HCl at infinite dilution = 426.16 S cm2 mol1
Molar conductivity of CH3COONa at infinite dilution = 91 S cm2 mol1
We need to find the molar conductivity of CH3COOH.
Step 1: Write the dissociation equation for CH3COOH.
CH3COOH ⇌ CH3COO- + H+
Step 2: Determine the number of ions produced by the dissociation of CH3COOH.
In the dissociation equation, one CH3COOH molecule produces one CH3COO- ion and one H+ ion. So, the total number of ions produced is 2.
Step 3: Apply Kohlrausch's law to calculate the molar conductivity of CH3COOH.
According to Kohlrausch's law, the molar conductivity of CH3COOH can be calculated using the equation:
Λ(CH3COOH) = Λ(CH3COO-) + Λ(H+)
Given that the molar conductivity of NaCl is 126.45 S cm2 mol1 and it dissociates into Na+ and Cl- ions, we can write:
Λ(NaCl) = Λ(Na+) + Λ(Cl-)
Similarly, the molar conductivity of HCl is 426.16 S cm2 mol1 and it dissociates into H+ and Cl- ions, so:
Λ(HCl) = Λ(H+) + Λ(Cl-)
Using the above equations and the given molar conductivities, we can rearrange the equation to solve for the molar conductivity of CH3COOH:
Λ(CH3COOH) = Λ(NaCl) - Λ(Cl-) + Λ(HCl) - Λ(H+)
Λ(CH3COOH) = 126.45 - 91 + 426.16 - 426.16
Simplifying the equation gives:
Λ(CH3COOH) = 201.28 S cm2 mol1
Therefore, the molar conductivity of CH3COOH at infinite dilution is 201.28 S cm2 mol1.
Hence, option 'a' is the correct answer.
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The molar conductivity of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91 S cm2 mol–1 respectively. The molar conductivity of CH3COOH at infinite dilution is:a)201.28 S cm2 mol–1b)698.28 S cm2 mol–1c)390.71 S cm2 mol–1d)540.48 S cm2 mol–1Correct answer is option 'C'. Can you explain this answer?
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The molar conductivity of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91 S cm2 mol–1 respectively. The molar conductivity of CH3COOH at infinite dilution is:a)201.28 S cm2 mol–1b)698.28 S cm2 mol–1c)390.71 S cm2 mol–1d)540.48 S cm2 mol–1Correct answer is option 'C'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about The molar conductivity of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91 S cm2 mol–1 respectively. The molar conductivity of CH3COOH at infinite dilution is:a)201.28 S cm2 mol–1b)698.28 S cm2 mol–1c)390.71 S cm2 mol–1d)540.48 S cm2 mol–1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The molar conductivity of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91 S cm2 mol–1 respectively. The molar conductivity of CH3COOH at infinite dilution is:a)201.28 S cm2 mol–1b)698.28 S cm2 mol–1c)390.71 S cm2 mol–1d)540.48 S cm2 mol–1Correct answer is option 'C'. Can you explain this answer?.
The molar conductivity of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91 S cm2 mol–1 respectively. The molar conductivity of CH3COOH at infinite dilution is:a)201.28 S cm2 mol–1b)698.28 S cm2 mol–1c)390.71 S cm2 mol–1d)540.48 S cm2 mol–1Correct answer is option 'C'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about The molar conductivity of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91 S cm2 mol–1 respectively. The molar conductivity of CH3COOH at infinite dilution is:a)201.28 S cm2 mol–1b)698.28 S cm2 mol–1c)390.71 S cm2 mol–1d)540.48 S cm2 mol–1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The molar conductivity of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91 S cm2 mol–1 respectively. The molar conductivity of CH3COOH at infinite dilution is:a)201.28 S cm2 mol–1b)698.28 S cm2 mol–1c)390.71 S cm2 mol–1d)540.48 S cm2 mol–1Correct answer is option 'C'. Can you explain this answer?.
Solutions for The molar conductivity of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91 S cm2 mol–1 respectively. The molar conductivity of CH3COOH at infinite dilution is:a)201.28 S cm2 mol–1b)698.28 S cm2 mol–1c)390.71 S cm2 mol–1d)540.48 S cm2 mol–1Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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The molar conductivity of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91 S cm2 mol–1 respectively. The molar conductivity of CH3COOH at infinite dilution is:a)201.28 S cm2 mol–1b)698.28 S cm2 mol–1c)390.71 S cm2 mol–1d)540.48 S cm2 mol–1Correct answer is option 'C'. Can you explain this answer?, a detailed solution for The molar conductivity of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91 S cm2 mol–1 respectively. The molar conductivity of CH3COOH at infinite dilution is:a)201.28 S cm2 mol–1b)698.28 S cm2 mol–1c)390.71 S cm2 mol–1d)540.48 S cm2 mol–1Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of The molar conductivity of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91 S cm2 mol–1 respectively. The molar conductivity of CH3COOH at infinite dilution is:a)201.28 S cm2 mol–1b)698.28 S cm2 mol–1c)390.71 S cm2 mol–1d)540.48 S cm2 mol–1Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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