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The molar conductivity of acetic acid at infinite dilution is 387per ohm square centimetre per mole . At the same temperature but at a concentration of one mole in 1000 litres it is 55 per ohm square centimetre per mole. What is the percentage dissociation of 0.001M acetic acid?
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Most Upvoted Answer
The molar conductivity of acetic acid at infinite dilution is 387per o...
The percentage of dissociation of acetic acid is 14.2%
Explanation:
Given data:
molar conductivity of acetic acid at infinite dilution = 387 ohm^-1 cm^2 / mol.
=molar conductance of solution at particular or specific concentration
The temperature is maintained constant.
concentration of 1 mole in 1000 litres = 55 ohm^-1 cm^2 / mol
= Molar conduction at infinite dilution
To find:
the percentage dissociation of acetic acid
Solution:
We know that
Aplha = a = degree of dissociation
= molar conductance of solution at particular or specific concentration/
Molar conduction at infinite dilution
=55/387
= 0.142
Percentage of degree of dissociation= 0.142*100
=14.2%
Hence,
The percentage of dissociation of acetic acid is 14.2%

Community Answer
The molar conductivity of acetic acid at infinite dilution is 387per o...
The percentage dissociation of acetic acid can be determined using the molar conductivity values at infinite dilution and at a specific concentration. Let's break down the problem step by step:

Given data:
Molar conductivity at infinite dilution (Λ∞) = 387 per ohm square centimetre per mole
Molar conductivity at concentration 1 M/1000 L (Λc) = 55 per ohm square centimetre per mole

To calculate the percentage dissociation of acetic acid, we need to determine the degree of dissociation (α) at a concentration of 0.001 M. The molar conductivity at this concentration is not provided directly, but we can approximate it using the given data.

Step 1: Calculate the molar conductivity at 0.001 M concentration
We can use the relationship between molar conductivity and concentration:

Λc = Λ∞ - K√c

Where K is an empirical constant and c is the concentration. Rearranging the equation, we have:

K√c = Λ∞ - Λc

Squaring both sides:

K^2c = Λ∞^2 + Λc^2 - 2Λ∞Λc

To calculate the molar conductivity at 0.001 M concentration, we substitute the given values:

K^2(0.001) = (387)^2 + (55)^2 - 2(387)(55)

Solving this equation will give us the value of K^2. Let's assume it to be X for simplicity.

Step 2: Calculate the percentage dissociation
The degree of dissociation (α) is defined as the ratio of the molar conductivity at a specific concentration to the molar conductivity at infinite dilution:

α = Λc / Λ∞

Substituting the given values:

α = 55 / 387

The percentage dissociation can be calculated by multiplying the degree of dissociation by 100:

Percentage dissociation = α * 100

Simplifying this equation will give us the final answer.

By following these steps and using the given data, you can calculate the percentage dissociation of acetic acid at a concentration of 0.001 M.
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The molar conductivity of acetic acid at infinite dilution is 387per ohm square centimetre per mole . At the same temperature but at a concentration of one mole in 1000 litres it is 55 per ohm square centimetre per mole. What is the percentage dissociation of 0.001M acetic acid?
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