The values of `^^_(m)^(oo)` for KCl and `KNO_(3)` are 149.86 and `154.96Omega^(-1)cm^(2)"mol"^(-1)`. Also `lambda_(Cl)^(oo)` is 71.44 `ohm^(-1)cm^(2) "mol"^(-1)` . The value of `lambda_(NO_(3)^(-))^(oo)` is

Options

A. 76.54 `ohm^(-1) cm^(2)"mol"^(-1)`
B. `133-08 ohm^(-1)cm^(2)"mol"^(-1)`
C. 37.7 `ohm^(-1)cm^(2)"mol"^(-1)`
D. unpredictable.

Explanation

To find the limiting molar conductivity of the nitrate ion \((\lambda_{NO_3^-}^{\infty})\), we can use the relationship between the molar conductivities of the salts and their constituent ions. The molar conductivity of a salt at infinite dilution is the sum of the molar conductivities of its cation and anion.

### Step-by-Step Solution:

1. **Write the equation for KCl:**
\[ \lambda_{KCl}^{\infty} = \lambda_{K^+}^{\infty} + \lambda_{Cl^-}^{\infty} \]
Given:
\[ \lambda_{KCl}^{\infty} = 149.86 \, \Omega^{-1} \, cm^2 \, mol^{-1} \]
We need to find \(\lambda_{K^+}^{\infty}\) and \(\lambda_{Cl^-}^{\infty}\).

2. **Write the equation for KNO3:**
\[ \lambda_{KNO_3}^{\infty} = \lambda_{K^+}^{\infty} + \lambda_{NO_3^-}^{\infty} \]
Given:
\[ \lambda_{KNO_3}^{\infty} = 154.96 \, \Omega^{-1} \, cm^2 \, mol^{-1} \]

3. **Substitute the known values:**
We know \(\lambda_{Cl^-}^{\infty} = 71.44 \, \Omega^{-1} \, cm^2 \, mol^{-1}\).
From the KCl equation:
\[ 149.86 = \lambda_{K^+}^{\infty} + 71.44 \]

4. **Solve for \(\lambda_{K^+}^{\infty}\):**
\[ \lambda_{K^+}^{\infty} = 149.86 - 71.44 = 78.42 \, \Omega^{-1} \, cm^2 \, mol^{-1} \]

5. **Substitute \(\lambda_{K^+}^{\infty}\) into the KNO3 equation:**
\[ 154.96 = 78.42 + \lambda_{NO_3^-}^{\infty} \]

6. **Solve for \(\lambda_{NO_3^-}^{\infty}\):**
\[ \lambda_{NO_3^-}^{\infty} = 154.96 - 78.42 = 76.54 \, \Omega^{-1} \, cm^2 \, mol^{-1} \]

### Final Answer:
\[ \lambda_{NO_3^-}^{\infty} = 76.54 \, \Omega^{-1} \, cm^2 \, mol^{-1} \]