The limiting molar conductivity for KCl, NaBr, and NaCl is 100, 120, and 130 S cm^-2 mol^-1. Calculate the limiting molar conductivity for KBr.

To calculate the limiting molar conductivity for KBr, we can use the Kohlrausch's law of independent migration of ions:

Λ(KBr) = Λ(K⁺) + Λ(Br⁻)

Given:
Λ(KCl) = 100 S cm^-2 mol^-1
Λ(NaBr) = 120 S cm^-2 mol^-1
Λ(NaCl) = 130 S cm^-2 mol^-1

We know that Λ(Na⁺) = Λ(Cl⁻) = 50 S cm^-2 mol^-1 (since NaCl has a limiting molar conductivity of 130 S cm^-2 mol^-1)

Therefore, Λ(K⁺) = Λ(KCl) - Λ(Cl⁻) = 100 - 50 = 50 S cm^-2 mol^-1
And, Λ(Br⁻) = Λ(NaBr) - Λ(Na⁺) = 120 - 50 = 70 S cm^-2 mol^-1

Now, we can calculate the limiting molar conductivity for KBr:
Λ(KBr) = Λ(K⁺) + Λ(Br⁻) = 50 + 70 = 120 S cm^-2 mol^-1

Therefore, the limiting molar conductivity for KBr is 120 S cm^-2 mol^-1.

Kohlrausch's Law of Independent Migration of Ions states that the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions. This law can be expressed as:

$$\Lambda^{\circ}_{cation} + \Lambda^{\circ}_{anion} = \Lambda^{\circ}_{electrolyte}$$

Where:

- \(\Lambda^{\circ}_{cation}\) is the limiting molar conductivity of the cation
- \(\Lambda^{\circ}_{anion}\) is the limiting molar conductivity of the anion
- \(\Lambda^{\circ}_{electrolyte}\) is the limiting molar conductivity of the electrolyte

For the given electrolytes, we have:

- KCl: \(\Lambda^{\circ}_{KCl} = 100\) S cm^-2 mol^-1
- NaBr: \(\Lambda^{\circ}_{NaBr} = 120\) S cm^-2 mol^-1
- NaCl: \(\Lambda^{\circ}_{NaCl} = 130\) S cm^-2 mol^-1
- KBr: \(\Lambda^{\circ}_{KBr} = \Lambda^{\circ}_{K^+} + \Lambda^{\circ}_{Br^-}\)

Subtracting \(\Lambda^{\circ}_{NaCl}\) from both sides of the equation for \(\Lambda^{\circ}_{KBr}\), we get:

$$\Lambda^{\circ}_{KBr} - \Lambda^{\circ}_{NaCl} = \Lambda^{\circ}_{K^+} + \Lambda^{\circ}_{Br^-} - \Lambda^{\circ}_{Na^+} - \Lambda^{\circ}_{Cl^-}$$

Since \(\Lambda^{\circ}_{Na^+} = \Lambda^{\circ}_{K^+}\) and \(\Lambda^{\circ}_{Cl^-} = \Lambda^{\circ}_{Br^-}\), the above equation simplifies to:

$$\Lambda^{\circ}_{KBr} = \Lambda^{\circ}_{NaCl} + \Lambda^{\circ}_{K^+}-\Lambda^{\circ}_{Na^+}$$

Substituting the given values, we get:

$$\Lambda^{\circ}_{KBr} = 130 + 100 - 120 = \boxed{110}\) S cm^-2 mol^-1