The value of `^^_(m)^(infty)` for KCl and `KNO_(3)` are `149.86` and `154.96 Omega^(-1) cm^(2) mol^(-1)`
Also `lambda_(Cl^(-))^(infty)` is `71.44 " ohm"^(-1) cm^(2)mol^(-1)` The value of `lambda_(NO_(3)^(-))^(infty)` 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 value of \( \lambda_{NO_3^-}^{\infty} \) using the given data for KCl and KNO₃, we will follow these steps:

### Step 1: Write down the given values
We have the following values:
- \( \Lambda_{KCl}^{\infty} = 149.86 \, \Omega^{-1} \, cm^2 \, mol^{-1} \)
- \( \Lambda_{KNO_3}^{\infty} = 154.96 \, \Omega^{-1} \, cm^2 \, mol^{-1} \)
- \( \lambda_{Cl^-}^{\infty} = 71.44 \, \Omega^{-1} \, cm^2 \, mol^{-1} \)

### Step 2: Write the equations for the conductance at infinite dilution
For KCl, the total conductance at infinite dilution can be expressed as:
\[ \Lambda_{KCl}^{\infty} = \lambda_{K^+}^{\infty} + \lambda_{Cl^-}^{\infty} \]
For KNO₃, the total conductance at infinite dilution can be expressed as:
\[ \Lambda_{KNO_3}^{\infty} = \lambda_{K^+}^{\infty} + \lambda_{NO_3^-}^{\infty} \]

### Step 3: Set up the equations
From the above equations, we can set up the following:
1. \( \lambda_{K^+}^{\infty} + \lambda_{Cl^-}^{\infty} = 149.86 \) (Equation 1)
2. \( \lambda_{K^+}^{\infty} + \lambda_{NO_3^-}^{\infty} = 154.96 \) (Equation 2)

### Step 4: Subtract Equation 1 from Equation 2
To eliminate \( \lambda_{K^+}^{\infty} \), we subtract Equation 1 from Equation 2:
\[ (\lambda_{K^+}^{\infty} + \lambda_{NO_3^-}^{\infty}) - (\lambda_{K^+}^{\infty} + \lambda_{Cl^-}^{\infty}) = 154.96 - 149.86 \]
This simplifies to:
\[ \lambda_{NO_3^-}^{\infty} - \lambda_{Cl^-}^{\infty} = 5.10 \]

### Step 5: Substitute the value of \( \lambda_{Cl^-}^{\infty} \)
Now, we can substitute the value of \( \lambda_{Cl^-}^{\infty} \):
\[ \lambda_{NO_3^-}^{\infty} - 71.44 = 5.10 \]

### Step 6: Solve for \( \lambda_{NO_3^-}^{\infty} \)
Adding \( 71.44 \) to both sides gives:
\[ \lambda_{NO_3^-}^{\infty} = 5.10 + 71.44 = 76.54 \, \Omega^{-1} \, cm^2 \, mol^{-1} \]

### Final Answer
Thus, the value of \( \lambda_{NO_3^-}^{\infty} \) is:
\[ \lambda_{NO_3^-}^{\infty} = 76.54 \, \Omega^{-1} \, cm^2 \, mol^{-1} \]

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