The conductivity of 0.2M methanoic acid is `8Sm^(-1)`. Then, degree of dissociation for methanoic acid is: [Given : `lambda_((H^(+)))^(@)=350 Scm^(2)"mol"^(-1),lambda_(HCOO^(-))^(@)=50 Scm^(2) "mol"^(-1)`]

Options

A. `10^(-1)`
B. `10^(-3)`
C. `10^(-5)`
D. `10^(-7)`

Answer

To find the degree of dissociation (α) of methanoic acid, we can follow these steps:

### Step 1: Understand the given data
We are given:
- Conductivity (κ) of 0.2 M methanoic acid = 0.8 S/m
- Molar conductivity of H⁺ (λ(H⁺)) at infinite dilution = 350 S cm²/mol
- Molar conductivity of HCOO⁻ (λ(HCOO⁻)) at infinite dilution = 50 S cm²/mol

### Step 2: Convert conductivity to appropriate units
Convert the conductivity from S/m to S/cm:
\[ \kappa = 0.8 \, \text{S/m} = \frac{0.8}{100} \, \text{S/cm} = 0.008 \, \text{S/cm} \]

### Step 3: Calculate the molar conductivity at infinite dilution
The molar conductivity (Λ) of methanoic acid at infinite dilution is the sum of the molar conductivities of its ions:
\[ \Lambda_0 = \lambda(H^+) + \lambda(HCOO^-) = 350 \, \text{S cm}^2/\text{mol} + 50 \, \text{S cm}^2/\text{mol} = 400 \, \text{S cm}^2/\text{mol} \]

### Step 4: Calculate the molar conductivity at the given concentration
The molar conductivity (Λ) at a specific concentration can be calculated using the formula:
\[ \Lambda = \frac{\kappa}{C} \times 1000 \]
where C is the concentration in mol/L (0.2 M in this case):
\[ \Lambda = \frac{0.008 \, \text{S/cm}}{0.2 \, \text{mol/L}} \times 1000 = \frac{0.008}{0.2} \times 1000 = 40 \, \text{S cm}^2/\text{mol} \]

### Step 5: Calculate the degree of dissociation (α)
The degree of dissociation (α) can be calculated using the formula:
\[ \alpha = \frac{\Lambda}{\Lambda_0} \]
Substituting the values we calculated:
\[ \alpha = \frac{40 \, \text{S cm}^2/\text{mol}}{400 \, \text{S cm}^2/\text{mol}} = \frac{1}{10} = 0.1 \]

### Final Answer
The degree of dissociation (α) for methanoic acid is **0.1** or **10%**.

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