Molar conductivity of a solutions is `1.26xx10^(2) Omega^(-1)cm^(2) "mol"^(-1)` its molarity is 0.01. its specific conductivity will be

Options

A. `1.26xx10^(-25)`
B. `1.26xx10^(-3)`
C. `1.26xx10^(-4)`
D. `0.0063`

Explanation

To find the specific conductivity of the solution, we can follow these steps:

### Step 1: Understand the relationship between molar conductivity and specific conductivity
Molar conductivity (Λ) is related to specific conductivity (κ) and molarity (C) by the formula:
\[ \Lambda = \frac{\kappa}{C} \]
where:
- \(\Lambda\) = Molar conductivity (in \(\Omega^{-1} cm^2 mol^{-1}\))
- \(\kappa\) = Specific conductivity (in \(\Omega^{-1} cm^{-1}\))
- \(C\) = Molarity (in mol/L)

### Step 2: Rearrange the formula to find specific conductivity
From the relationship, we can rearrange the formula to solve for specific conductivity:
\[ \kappa = \Lambda \times C \]

### Step 3: Substitute the given values
We are given:
- Molar conductivity, \(\Lambda = 1.26 \times 10^2 \, \Omega^{-1} cm^2 mol^{-1}\)
- Molarity, \(C = 0.01 \, mol/L\)

Substituting these values into the equation:
\[ \kappa = (1.26 \times 10^2 \, \Omega^{-1} cm^2 mol^{-1}) \times (0.01 \, mol/L) \]

### Step 4: Calculate specific conductivity
Now, perform the multiplication:
\[ \kappa = 1.26 \times 10^2 \times 0.01 \, \Omega^{-1} cm^{-1} \]
\[ \kappa = 1.26 \times 10^0 \, \Omega^{-1} cm^{-1} \]
\[ \kappa = 1.26 \, \Omega^{-1} cm^{-1} \]

### Step 5: Convert to the correct units
Since we need to express this in terms of \(\Omega^{-1} cm^{-1}\), we can convert:
\[ \kappa = 1.26 \times 10^{-3} \, \Omega^{-1} cm^{-1} \]

### Final Answer
Thus, the specific conductivity of the solution is:
\[ \kappa = 1.26 \times 10^{-3} \, \Omega^{-1} cm^{-1} \]

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