Resistance of a 0.1 M KCl solution in a conductance cell is 300 ohm and specific conductance of `"0.1 M KCl"` is `1.33xx10^(-2)" ohm"^(-1)"cm"^(-1)`. The resistance of 0.1 M NaCl solution in the same cell is 400 ohm. The equivalent conductance of the 0.1 M NaCl solution `("in ohm"^(-1)"cm"^(2)"/gmeq.")` is

Answer

To solve the problem, we need to find the equivalent conductance of a 0.1 M NaCl solution using the provided data. Let's go through the steps systematically.

### Step 1: Understand the given data
- Resistance of 0.1 M KCl solution (R_KCl) = 300 ohm
- Specific conductance of 0.1 M KCl (κ_KCl) = 1.33 × 10^(-2) ohm^(-1) cm^(-1)
- Resistance of 0.1 M NaCl solution (R_NaCl) = 400 ohm

### Step 2: Calculate the cell constant (x)
The cell constant (x) can be calculated using the specific conductance of KCl and its resistance.

Using the formula:
\[ κ = C \cdot x \]
Where:
- κ = specific conductance
- C = conductance = \( \frac{1}{R} \)

For KCl:
\[ C_{KCl} = \frac{1}{R_{KCl}} = \frac{1}{300} \, \text{S} \]

Now substituting into the equation:
\[ 1.33 \times 10^{-2} = \left(\frac{1}{300}\right) \cdot x \]

Rearranging to find x:
\[ x = 1.33 \times 10^{-2} \times 300 \]
\[ x = 3.99 \approx 4 \, \text{cm}^{-1} \]

### Step 3: Calculate the conductance of 0.1 M NaCl
Using the resistance of the NaCl solution:
\[ C_{NaCl} = \frac{1}{R_{NaCl}} = \frac{1}{400} \, \text{S} \]

### Step 4: Calculate the specific conductance of NaCl
Using the cell constant:
\[ κ_{NaCl} = C_{NaCl} \cdot x \]
Substituting the values:
\[ κ_{NaCl} = \left(\frac{1}{400}\right) \cdot 4 \]
\[ κ_{NaCl} = \frac{4}{400} = 0.01 \, \text{ohm}^{-1} \text{cm}^{-1} \]

### Step 5: Calculate the equivalent conductance (λ)
The equivalent conductance (λ) can be calculated using the formula:
\[ λ = \frac{κ \cdot 1000}{N} \]
Where:
- N = normality = molarity × n factor
- For NaCl, n factor = 1, thus N = 0.1

Substituting the values:
\[ λ = \frac{0.01 \cdot 1000}{0.1} \]
\[ λ = \frac{10}{0.1} = 100 \, \text{ohm}^{-1} \text{cm}^2/\text{gmeq} \]

### Final Answer
The equivalent conductance of the 0.1 M NaCl solution is:
\[ \boxed{100 \, \text{ohm}^{-1} \text{cm}^2/\text{gmeq}} \]