Question

The equivalent conductance of 0.1N acetic acid is 5cm2 ohm-1 eq-1 at 298K while lambda infty is 390cm2/ohm/eq. The degree of dissociation of 0.1N acetic acid will be

Answer 1

Answer:

0.013. kindly go through the explanation

Answer 2

Given

• Equivalent conductance of 0.1 N acetic acid $\Lambda_{eq} = 5 cm^2 ohm^{-1} eq^{-1}$

• $\Lambda_{\infty} = 390$

To Find

The degree of dissociation of 0.1 N acetic acid.

Formulas

$\Lambda_{eq} = n-factor×\Lambda_{molar}$

Degree of dissociation of a weak acid = $\frac{ \Lambda_{molar} }{ \lambda_{ \infty}}$

Solutions

Let us first find $\Lambda_{molar}$

$\Lambda_{eq} = (n-factor)×\Lambda_{molar}$

n-factor of $CH_3COOH$ is 1.

Hence, $\Lambda_{eq} = \Lambda_{molar}$

=> $5 = \Lambda_{molar}$

Now, degree of dissociation = $\frac{ \Lambda_{molar} }{ \lambda_{ \infty}}$

= $\frac{ 5 }{ 390}$

= 0.013

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