Question

The limiting molar conductivity of hcl, ch3coona and nacl are respectively 425,90 and 125 mho cm^2 mol^-1 at 25°c the molar conductivity of 0.1m ch3cooh solution is 7.8 mho cm^2 mol^-1 at the same temperature the degree of dissociation of o.1m acetic acid solution at the same temperature is

Answer 1

Answer:

0.02

First find limiting molar conductivity of ch3cooh

^° ch3cooh = ^°(ch3coona+hcl-nacl)

90+425-125

=390

Degree of dissociation = ^/^°

7.8/390 = 0.02

Answer 2

The degree of dissociation of acetic acid solution is 0.02

Explanation:

To calculate the equivalent conductance of $BaSO_4$ at infinite dilution, we will apply Kohlrausch Law.

This law states that the sum of the limiting ionic conductivities of the cation and the anion each multiplied with the number of ions that are present in one formula unit of the electrolyte.

The equation used to calculate the equivalent conductance of $CH_3COOH$ at infinite dilution is:

$\Lambda^o_{CH_3COOH}=\Lambda^o_{CH_3COONa}+\Lambda^o_{HCl}-\Lambda^o_{NaCl}$

We are given:

$\Lambda^o_{CH_3COONa}=90\Omega ^{-1}cm^2mol^{-1}\\\Lambda^o_{HCl}=425\Omega ^{-1}cm^2mol^{-1}\\Lambda^o_{NaCl}=125\Omega ^{-1}cm^2mol^{-1}$

Putting values in above equation, we get:

$\Lambda^o_{CH_3COOH}=[90+425-125]\\\\\Lambda^o_{CH_3COOH}=390\Omega ^{-1}cm^2mol^{-1}$

To calculate the degree of dissociation, we use the equation:

$\alpha=\frac{\Lambda _m}{\Lambda^o}$

We are given:

$\Lambda_m=7.8\Omega ^{-1}cm^2mol^{-1}\\\\\Lambda^o=390\Omega ^{-1}cm^2mol^{-1}$

Putting values in above equation, we get:

$\alpha =\frac{7.8}{390}=0.02$

Learn more about Kohlrausch law:

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