The dissociation constant of a weak acid is `1.6xx10^(-5)` and the molar conductivity at infinite dilution is `380xx10^(-4)` S`m^2mol^(-1)` . If the cell constant is 0.01`m^(-1)` then conductace of 0.1M acid solution is :

Options

A. `1.52xx10^(-5)`S
B. 1.52 S
C. `1.52xx10^(-3)`S
D. `1.52xx10^(-4)`S

Answer

To find the conductance of a 0.1 M weak acid solution given the dissociation constant and molar conductivity at infinite dilution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Dissociation constant (Kα) = \(1.6 \times 10^{-5}\)
- Molar conductivity at infinite dilution (Λm) = \(380 \times 10^{-4} \, \text{S m}^2 \text{mol}^{-1}\)
- Cell constant (G*) = \(0.01 \, \text{m}^{-1}\)
- Concentration (C) = \(0.1 \, \text{mol m}^{-3}\)

2. **Use the Relationship for Dissociation Constant:**
The dissociation constant for a weak acid can be expressed as:
\[ Kα = \frac{C \cdot α^2}{1 - α} \]
where \(α\) is the degree of dissociation.

3. **Substitute Known Values:**
Substitute the known values into the equation:
\[ 1.6 \times 10^{-5} = \frac{0.1 \cdot α^2}{1 - α} \]

4. **Assume Small α:**
Since \(Kα\) is small, we can assume that \(α\) is small enough that \(1 - α \approx 1\):
\[ 1.6 \times 10^{-5} \approx 0.1 \cdot α^2 \]

5. **Solve for α:**
Rearranging gives:
\[ α^2 = \frac{1.6 \times 10^{-5}}{0.1} = 1.6 \times 10^{-4} \]
Taking the square root:
\[ α = \sqrt{1.6 \times 10^{-4}} = 0.04 \]

6. **Calculate Molar Conductivity (Λ) of the Acid Solution:**
The molar conductivity (Λ) of the solution can be calculated using:
\[ Λ = α \cdot Λ_m \]
Substituting the values:
\[ Λ = 0.04 \cdot (380 \times 10^{-4}) = 15.2 \times 10^{-4} \, \text{S m}^2 \text{mol}^{-1} \]

7. **Convert Molar Conductivity to Conductivity (K):**
The conductivity (K) can be calculated as:
\[ K = \frac{Λ}{C} \]
Substituting the values:
\[ K = \frac{15.2 \times 10^{-4}}{0.1} = 15.2 \times 10^{-3} \, \text{S m}^{-1} \]

8. **Calculate Conductance (G):**
Finally, the conductance (G) can be calculated using the cell constant (G*):
\[ G = K \cdot G^* \]
Substituting the values:
\[ G = 15.2 \times 10^{-3} \cdot 0.01 = 1.52 \, \text{S} \]

### Final Answer:
The conductance of the 0.1 M acid solution is **1.52 S**.