Conductance of `0.1` M KCl (conductiviy = `X ohm^(-1) cm^(-1)`) filled in a conductivity cell is Y `ohm^(-1)` If the conductance of 0.1 M NaOH filled in the same cell is Z `ohm^(-1)` the molar conductance of NaOH will be

Options

A. `10^(3)(XZ)/Y`
B. `10^(4) (XZ)/Y`
C. `10(XZ)/Y`
D. `0.1(XZ)/Y`

Explanation

To find the molar conductance of NaOH using the information given about KCl and NaOH, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between conductivity, conductance, and cell constant:**
\[ \text{Conductivity} (k) = \text{Conductance} (G) \times \text{Cell Constant} (C) \]
Here, the conductivity of KCl is given as \( x \, \text{ohm}^{-1} \text{cm}^{-1} \) and the conductance is given as \( Y \, \text{ohm}^{-1} \).

2. **Calculate the cell constant (C) using KCl data:**
\[ C = \frac{k}{G} = \frac{x}{Y} \]

3. **Determine the conductivity of NaOH using its conductance:**
The conductance of 0.1 M NaOH is given as \( Z \, \text{ohm}^{-1} \). Using the same cell constant, we can express the conductivity of NaOH as:
\[ k_{\text{NaOH}} = Z \times C = Z \times \frac{x}{Y} \]

4. **Use the formula for molar conductance (Λ):**
Molar conductance is given by:
\[ \Lambda = \frac{k \times 1000}{\text{Molarity}} \]
For NaOH, the molarity is \( 0.1 \, \text{M} \):
\[ \Lambda_{\text{NaOH}} = \frac{k_{\text{NaOH}} \times 1000}{0.1} \]

5. **Substituting the expression for \( k_{\text{NaOH}} \):**
\[ \Lambda_{\text{NaOH}} = \frac{(Z \times \frac{x}{Y}) \times 1000}{0.1} \]

6. **Simplifying the expression:**
\[ \Lambda_{\text{NaOH}} = \frac{Z \times x \times 1000}{0.1 \times Y} = \frac{Z \times x \times 10000}{Y} \]

### Final Expression:
Thus, the molar conductance of NaOH is given by:
\[ \Lambda_{\text{NaOH}} = \frac{Z \times x \times 10000}{Y} \]