What will be the molar conductance `'Lamda'`, if resistivity is 'x' for 0.1 N `H_(2)SO_(4)` solution

Options

A. `Lamda= (x xx 1000)/(0.1)`
B. `Lamda = (2 xx 1000)/(x xx 0.1)`
C. `Lamda = (x xx 1000)/(0.5)`
D. `Lamda = (0.5)/(1000 x)`

Answer

To find the molar conductance (Λ) of a 0.1 N H₂SO₄ solution given its resistivity (x), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between Normality and Molarity**:
- The normality (N) of a solution is related to its molarity (M) by the equation:
\[ N = n \times M \]
- Here, \( n \) is the number of replaceable hydrogen ions per molecule of the acid. For sulfuric acid (H₂SO₄), \( n = 2 \) because it can donate two protons (H⁺ ions).

2. **Calculate Molarity from Normality**:
- Given that the normality of the H₂SO₄ solution is 0.1 N, we can find the molarity (M):
\[ M = \frac{N}{n} = \frac{0.1}{2} = 0.05 \, \text{M} \]

3. **Relate Conductance to Resistivity**:
- The specific conductance (κ) is the reciprocal of resistivity (ρ):
\[ κ = \frac{1}{ρ} = \frac{1}{x} \]

4. **Use the formula for Molar Conductance**:
- Molar conductance (Λ) is given by the formula:
\[ Λ = \frac{κ \times 1000}{M} \]
- Substituting the values we have:
\[ Λ = \frac{\left(\frac{1}{x}\right) \times 1000}{0.05} \]

5. **Simplify the expression**:
- This simplifies to:
\[ Λ = \frac{1000}{0.05 \times x} = \frac{1000 \times 20}{x} = \frac{20000}{x} \]

6. **Final Expression for Molar Conductance**:
- Thus, the final expression for the molar conductance of the 0.1 N H₂SO₄ solution is:
\[ Λ = \frac{20000}{x} \]

### Conclusion:
The molar conductance (Λ) of a 0.1 N H₂SO₄ solution with resistivity x is given by:
\[ Λ = \frac{20000}{x} \]