The specific conductance (`K`) of an electrolyte of `0.1` N concentration is related to equivalent conductance `(wedge_(aq))` by the following formula.

Options

A. `wedge_(aq)= K`
B. `wedge_(aq)= 10K`
C. `wedge_(aq)= 100K`
D. `wedge_(aq)= 10000K`

Answer

To solve the problem, we need to relate the specific conductance (K) of an electrolyte to its equivalent conductance (Λ). The relationship can be expressed using the formula:

\[ \Lambda = \frac{K \times 1000}{N} \]

Where:
- \( \Lambda \) is the equivalent conductance,
- \( K \) is the specific conductance,
- \( N \) is the normality of the solution.

Given that the normality \( N \) is \( 0.1 \) N, we can substitute this value into the formula.

### Step 1: Substitute the values into the formula
We have:
\[ \Lambda = \frac{K \times 1000}{0.1} \]

### Step 2: Simplify the equation
To simplify, we can multiply both sides by \( 0.1 \):
\[ \Lambda \times 0.1 = K \times 1000 \]

Now, dividing both sides by \( 0.1 \):
\[ \Lambda = K \times 10000 \]

### Step 3: Conclusion
Thus, the equivalent conductance \( \Lambda \) is related to the specific conductance \( K \) by the formula:
\[ \Lambda = K \times 10000 \]

### Final Answer
The equivalent conductance \( \Lambda \) of the electrolyte at \( 0.1 \) N concentration can be expressed as:
\[ \Lambda = K \times 10000 \]