A weak monobasic acid is 5% dissociated in 0.01 mol `dm^(-3)` solution. The limiting molar conductivity at infinite dilution is `4.00xx10^(-2) ohm^(-1) m^(2) mol^(-1)`. Calculate the conductivity of a 0.05 mol `dm^(-3)` solution of the acid.

Options

A. `8.94xx10^(-6) "ohm"^(-1) cm^(2) mol^(-1)`
B. `8.92xx10^(-4) "ohm"^(-1) cm^(2) mol^(-1)`
C. `4.46xx10^(-6) "ohm"^(-1) cm^(2) mol^(-1)`
D. `2.23xx10^(-5) "ohm"^(-1) cm^(2) mol^(-1)`

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant equations.

### Step 1: Identify the given data
- The degree of dissociation (α) is 5%, which can be expressed as:
\[ \alpha = \frac{5}{100} = 0.05 \]
- The concentration (C) of the weak acid is:
\[ C = 0.01 \, \text{mol/dm}^3 \]
- The limiting molar conductivity at infinite dilution (λm∞) is:
\[ \lambda_m^{\infty} = 4.00 \times 10^{-2} \, \Omega^{-1} \, \text{m}^2 \, \text{mol}^{-1} \]

### Step 2: Calculate the dissociation constant (Kα)
Using the formula for the dissociation constant:
\[ K_{\alpha} = C \alpha^2 \]
Substituting the values:
\[ K_{\alpha} = 0.01 \times (0.05)^2 = 0.01 \times 0.0025 = 2.5 \times 10^{-5} \]

### Step 3: Use the dissociation constant for a new concentration
Now, we will calculate the degree of dissociation (α) for a new concentration of 0.05 mol/dm³. The dissociation constant remains the same:
\[ K_{\alpha} = C \alpha^2 \]
Substituting the known values:
\[ 2.5 \times 10^{-5} = 0.05 \times \alpha^2 \]
Rearranging gives:
\[ \alpha^2 = \frac{2.5 \times 10^{-5}}{0.05} = 5.0 \times 10^{-4} \]
Taking the square root:
\[ \alpha = \sqrt{5.0 \times 10^{-4}} \approx 0.02236 \approx 0.0223 \]

### Step 4: Calculate the molar conductivity at the given concentration
Using the relationship between conductivity (κ), degree of dissociation (α), and molar conductivity at infinite dilution (λm∞):
\[ \lambda_m = \alpha \cdot \lambda_m^{\infty} \]
Substituting the values:
\[ \lambda_m = 0.0223 \cdot (4.00 \times 10^{-2}) = 8.92 \times 10^{-4} \, \Omega^{-1} \, \text{m}^2 \, \text{mol}^{-1} \]

### Step 5: Calculate the conductivity (κ)
The conductivity (κ) can be calculated using the formula:
\[ \kappa = C \cdot \lambda_m \]
Substituting the values:
\[ \kappa = 0.05 \cdot (8.92 \times 10^{-4}) = 4.46 \times 10^{-5} \, \Omega^{-1} \, \text{m}^{-1} \]

### Final Answer
The conductivity of a 0.05 mol/dm³ solution of the acid is:
\[ \kappa = 4.46 \times 10^{-5} \, \Omega^{-1} \, \text{m}^{-1} \]