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 (ρ) as 'x', we can follow these steps:

### Step 1: Understand the relationship between normality and molarity
Normality (N) is related to molarity (M) by the equation:
\[ N = M \times n \]
where \( n \) is the number of equivalents of the solute. For sulfuric acid (H₂SO₄), the \( n \) factor is 2 because it can donate 2 protons (H⁺ ions) per molecule.

### Step 2: Convert normality to molarity
Given that the normality of the solution is 0.1 N, we can find the molarity (M) as follows:
\[ M = \frac{N}{n} = \frac{0.1}{2} = 0.05 \, M \]

### Step 3: Relate resistivity to specific conductance
Resistivity (ρ) is related to specific conductance (κ) by the equation:
\[ κ = \frac{1}{ρ} \]
Since the resistivity is given as 'x', we have:
\[ κ = \frac{1}{x} \]

### Step 4: Calculate molar conductance
Molar conductance (Λ) can be calculated using the formula:
\[ Λ = κ \times \frac{1000}{M} \]
Substituting the values we have:
\[ Λ = \left(\frac{1}{x}\right) \times \frac{1000}{0.05} \]

### Step 5: Simplify the expression
Now we simplify the expression:
\[ Λ = \frac{1000}{0.05 \times x} \]
\[ Λ = \frac{1000 \times 20}{x} \]
\[ Λ = \frac{20000}{x} \]

### Final Answer
Thus, the molar conductance (Λ) for the 0.1 N H₂SO₄ solution is:
\[ Λ = \frac{20000}{x} \]

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