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Find the molar conductivity of acetic acid if its conductivity is given to be 0.00241 M. Also, if the value of \Lambda^0_mΛm0​ is given to be390.5 S cm2 mol−1, calculate its dissociation constant?
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Find the molar conductivity of acetic acid if its conductivity is give...
""""K is missing""" K=7. 89×10^-5
lamdaa m =(k×1000)÷M

alpha =lamda M÷lamda'o M


k=*C × alpha ^2)÷(1-- alpha)
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Find the molar conductivity of acetic acid if its conductivity is give...
Calculation of Molar Conductivity:

To find the molar conductivity of acetic acid, we need to use the formula:

Molar conductivity (\(\Lambda_m\)) = Conductivity (\(\kappa\)) / Concentration (C)

Given that the conductivity (\(\kappa\)) of acetic acid is 0.00241 M, we can substitute this value into the formula:

\(\Lambda_m\) = 0.00241 M / C

Calculation of Dissociation Constant:

The dissociation constant (\(K_a\)) can be calculated using the formula:

\(K_a\) = \(\frac{{\Lambda_m}}{{\Lambda_m^0}}\)

Given that the value of \(\Lambda_m^0\) is 390.5 S cm² mol⁻¹, we can substitute this value into the formula:

\(K_a\) = \(\frac{{0.00241 M}}{{390.5 S cm² mol⁻¹}}\)

Simplifying the equation, we get:

\(K_a\) = 6.18 x 10⁻⁶

Explanation:

Acetic acid (CH₃COOH) is a weak acid that partially dissociates in water to form acetate ions (CH₃COO⁻) and hydrogen ions (H⁺). The molar conductivity (\(\Lambda_m\)) of acetic acid is a measure of its ability to conduct electricity in solution and is related to the degree of dissociation.

The molar conductivity can be calculated by dividing the conductivity (\(\kappa\)) of acetic acid by its concentration (C). In this case, the conductivity is given as 0.00241 M.

The dissociation constant (\(K_a\)) represents the extent of dissociation of an acid in solution. It is calculated by dividing the molar conductivity (\(\Lambda_m\)) by the molar conductivity at infinite dilution (\(\Lambda_m^0\)). In this case, the value of \(\Lambda_m^0\) is given as 390.5 S cm² mol⁻¹.

By substituting the given values into the formula, we can calculate the dissociation constant (\(K_a\)) of acetic acid, which is found to be 6.18 x 10⁻⁶.

The dissociation constant is an important parameter in understanding the behavior of weak acids in solution. It provides information about the degree of dissociation, the strength of the acid, and the equilibrium between the dissociated and undissociated forms.
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Find the molar conductivity of acetic acid if its conductivity is given to be 0.00241 M. Also, if the value of \Lambda^0_mΛm0​ is given to be390.5 S cm2 mol−1, calculate its dissociation constant?
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Find the molar conductivity of acetic acid if its conductivity is given to be 0.00241 M. Also, if the value of \Lambda^0_mΛm0​ is given to be390.5 S cm2 mol−1, calculate its dissociation constant? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Find the molar conductivity of acetic acid if its conductivity is given to be 0.00241 M. Also, if the value of \Lambda^0_mΛm0​ is given to be390.5 S cm2 mol−1, calculate its dissociation constant? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find the molar conductivity of acetic acid if its conductivity is given to be 0.00241 M. Also, if the value of \Lambda^0_mΛm0​ is given to be390.5 S cm2 mol−1, calculate its dissociation constant?.
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