Conductance and conductivity of `0.1MKCl` solution filled in a conductivity cell are `"y ohm"^(–1) and "x ohm"^(–1)" cm"^( –1)` respectively. If the conductance of `"0.1 M NaOH"` filled in the same cell is `"z ohm"^(–1)` , molar conductance of `NaOH` will be :

Options

A. `0.1(xz)/(y)`
B. `10(xz)/(y)`
C. `10^(3)(xz)/(y)`
D. `10^(4)(xz)/(y)`

Explanation

To solve the problem, we need to find the molar conductance of a 0.1 M NaOH solution using the given conductance and conductivity of a 0.1 M KCl solution. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Definitions
- **Conductance (G)**: It is the reciprocal of resistance and is given in ohm\(^{-1}\).
- **Conductivity (κ)**: It is a measure of a solution's ability to conduct electricity and is given in ohm\(^{-1}\) cm\(^{-1}\).
- **Molar Conductance (Λ)**: It is defined as the conductivity of a solution multiplied by the volume of the solution per mole of solute, expressed in ohm\(^{-1}\) cm\(^2\) mol\(^{-1}\).

### Step 2: Write Down the Given Information
- Conductance of 0.1 M KCl: \( G = y \, \text{ohm}^{-1} \)
- Conductivity of 0.1 M KCl: \( κ = x \, \text{ohm}^{-1} \, \text{cm}^{-1} \)
- Conductance of 0.1 M NaOH: \( G_{NaOH} = z \, \text{ohm}^{-1} \)

### Step 3: Relate Conductance and Conductivity
The relationship between conductance (G), conductivity (κ), and the cell constant (L/A) is given by:
\[ G = κ \cdot \frac{A}{L} \]
Where \( \frac{A}{L} \) is the cell constant.

### Step 4: Calculate the Cell Constant
From the KCl solution, we can express the cell constant as:
\[ \frac{A}{L} = \frac{G}{κ} = \frac{y}{x} \]

### Step 5: Write the Conductance for NaOH
For the NaOH solution, we can write:
\[ G_{NaOH} = κ_{NaOH} \cdot \frac{A}{L} \]
Substituting the cell constant:
\[ z = κ_{NaOH} \cdot \frac{y}{x} \]

### Step 6: Solve for Conductivity of NaOH
Rearranging gives us:
\[ κ_{NaOH} = z \cdot \frac{x}{y} \]

### Step 7: Calculate Molar Conductance
The molar conductance (Λ) can be calculated using the formula:
\[ Λ = \frac{κ \cdot 1000}{C} \]
Where \( C \) is the concentration in mol/L (0.1 M for NaOH).

Substituting for \( κ_{NaOH} \):
\[ Λ_{NaOH} = \frac{z \cdot \frac{x}{y} \cdot 1000}{0.1} \]

### Step 8: Simplify the Expression
This simplifies to:
\[ Λ_{NaOH} = z \cdot \frac{x}{y} \cdot 10000 \]
Or:
\[ Λ_{NaOH} = z \cdot \frac{x}{y} \cdot 10^4 \]

### Final Answer
Thus, the molar conductance of NaOH is:
\[ Λ_{NaOH} = z \cdot \frac{x}{y} \cdot 10^4 \]

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