The limiting molar conductivities `overset(o)(wedge)` for NaCl, KBr and KCl are 126, 152 and `150 S cm^(2)mol^(-1)` respectively. The `overset(o)(wedge)` for NaBr is:

Options

A. `128Scm^(2) mol^(-1)`
B. `302Scm^(2)mol^(-1)`
C. `278Scm^(2mol^(-1)`
D. `176 S cm^(2)mol^(1)`

Explanation

To find the limiting molar conductivity (Λ°) of NaBr, we can use the relationship between the limiting molar conductivities of NaCl, KBr, and KCl. The formula we will use is based on the principle of additive conductivities for ions.

### Step-by-step Solution:

1. **Identify the given values:**
- Limiting molar conductivity of NaCl (Λ°(NaCl)) = 126 S cm² mol⁻¹
- Limiting molar conductivity of KBr (Λ°(KBr)) = 152 S cm² mol⁻¹
- Limiting molar conductivity of KCl (Λ°(KCl)) = 150 S cm² mol⁻¹
...