Limiting molar conductivity of NaBr is

Options

A. `Lambda_m^@NaBr = Lambda_m^@NaCl + Lambda_m^@KBr`
B. `Lambda_m^@NaBr = Lambda_m^@ NaCl + Lambda_m^@KBr - Lambda_m^@KCl`
C. `Lambda_m^@NaBr=Lambda_m^@NaOH + Lambda_m^@NaBr - Lambda_m^@NaCl`
D. `Lambda_m^@NaBr=Lambda_m^@NaCl-Lambda_m^@NaBr`

Answer

To find the limiting molar conductivity (λ₀) of sodium bromide (NaBr), we can use Kohlrausch's Law, which states that the molar conductivity of an ionic compound is equal to the sum of the molar conductivities of its constituent ions.

### Step-by-Step Solution:

1. **Understanding Kohlrausch's Law**:
According to Kohlrausch's Law, the limiting molar conductivity of an ionic compound (like NaBr) can be expressed as:
\[ \lambda_m^0(\text{NaBr}) = \lambda^0(\text{Na}^+) + \lambda^0(\text{Br}^-) \]

2. **Identify Known Values**:
We need the limiting molar conductivities of NaCl, KBr, and KCl to find the limiting molar conductivity of NaBr.
- Let’s denote:
- \(\lambda_m^0(\text{NaCl})\) = limiting molar conductivity of NaCl
- \(\lambda_m^0(\text{KBr})\) = limiting molar conductivity of KBr
- \(\lambda_m^0(\text{KCl})\) = limiting molar conductivity of KCl

3. **Using Relationships**:
We can express the limiting molar conductivity of NaBr in terms of the known values:
\[ \lambda_m^0(\text{NaBr}) = \lambda_m^0(\text{NaCl}) + \lambda_m^0(\text{KBr}) - \lambda_m^0(\text{KCl}) \]

4. **Substituting Values**:
Substitute the known values of the limiting molar conductivities of NaCl, KBr, and KCl into the equation to calculate \(\lambda_m^0(\text{NaBr})\).

5. **Final Expression**:
After substituting the values, you will arrive at the final expression for the limiting molar conductivity of NaBr.

### Conclusion:
Thus, the limiting molar conductivity of NaBr can be calculated using the relationship derived from Kohlrausch's Law.