Following limiting molar conductivities are given as

$$\begin{gathered} {\mathrm{\lambda }}_{\mathrm{m}}^{\circ }\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)=\mathrm{x} {\mathrm{Scm}}^2{\mathrm{mol}}^{-1} \\ {\mathrm{\lambda }}_{\mathrm{m}}^{\circ }\left({\mathrm{K}}_2{\mathrm{SO}}_4\right)=\mathrm{y} {\mathrm{Scm}}^2{\mathrm{mol}}^{-1} \\ {\mathrm{\lambda }}_{\mathrm{m}}^{\circ }\left({\mathrm{CH}}_3\mathrm{COOK}\right)=\mathrm{z} {\mathrm{Scm}}^2{\mathrm{mol}}^{-1} \\ {\mathrm{\lambda }}_{\mathrm{m}}^{\circ } (\mathrm{in} {\mathrm{Scm}}^2{\mathrm{mol}}^{-1} ) \mathrm{for} {\mathrm{CH}}_3\mathrm{COOH} \mathrm{willbe} \end{gathered}$$

We are given the following limiting molar conductivities (λ_m^°) for different compounds:

• λ_m^°(H₂SO₄) = x S·cm²·mol⁻¹
• λ_m^°(K₂SO₄) = y S·cm²·mol⁻¹
• λ_m^°(CH₃COOK) = z S·cm²·mol⁻¹

The task is to find the limiting molar conductivity (λ_m^°) of acetic acid (CH₃COOH).

Step 1: Understanding the dissociation of each compound

Let's start by writing the dissociation reactions of each compound:

1. Acetic acid: CH₃COOH → CH₃COO⁻ + H⁺ (Equation 1)
2. Sulfuric acid: H₂SO₄ → 2H⁺ + SO₄²⁻ (Equation 2)
3. Potassium sulfate: K₂SO₄ → 2K⁺ + SO₄²⁻ (Equation 3)
4. Potassium acetate: CH₃COOK → CH₃COO⁻ + K⁺ (Equation 4)

Step 2: Applying Kohlrausch's Law

Kohlrausch's law of independent migration of ions states that the limiting molar conductivity (λ_m^°) of an electrolyte can be expressed as the sum of the conductivities of its individual ions in their dissociated form:

For acetic acid (CH₃COOH):

λm°(CH3COOH) = λm°(CH3COO−) + λm°(H+)

Step 3: Deriving the formula for λ_m^°(CH₃COOH)

Now, using the dissociation equations, we can relate the limiting molar conductivities of various compounds:

For potassium acetate (CH₃COOK):

λm°(CH3COOK) = λm°(CH3COO−) + λm°(K+)

Similarly, for potassium sulfate (K₂SO₄):

λm°(K2SO4) = 2λm°(K+) + λm°(SO42−)

We can now apply Kohlrausch's law to the dissociation of CH₃COOH by substituting the appropriate values from the other compounds:

λm°(CH3COOH) = λm°(CH3COOK) + 2λm°(H2SO4) − 2λm°(K2SO4)

Step 4: Final Expression

Substitute the given values:

λm°(CH3COOH) = z + 2x − 2y

Thus, the limiting molar conductivity for acetic acid (CH₃COOH) is:

λm°(CH3COOH) = (x − y)/2 + z S·cm2·mol−1

Conclusion

The final result for the limiting molar conductivity of acetic acid (CH₃COOH) is:

λm°(CH3COOH) = (x − y)/2 + z