At`25^@C`, the specific conductance of 0.01 M alkaline earth metal chloride is `0.000158 "ohm"^(-1) cm^(-1)` Calculate the equivalent conductance.

Answer

To calculate the equivalent conductance of a 0.01 M alkaline earth metal chloride at 25°C, we can follow these steps:

### Step 1: Understand the relationship between molarity and normality
For alkaline earth metal chlorides (like MgCl₂, CaCl₂, etc.), the n-factor is 2 because each formula unit produces two moles of ions in solution.

### Step 2: Calculate the normality
Normality (N) is calculated using the formula:
\[ \text{Normality} = n \times \text{Molarity} \]
Here, n = 2 and Molarity = 0.01 M.
\[ \text{Normality} = 2 \times 0.01 \, \text{M} = 0.02 \, \text{N} \]

### Step 3: Use the formula for equivalent conductance
The equivalent conductance (Λ) can be calculated using the formula:
\[ \Lambda = \frac{1000 \times K}{N} \]
where:
- \( K \) is the specific conductance (given as \( 0.000158 \, \Omega^{-1} \, \text{cm}^{-1} \))
- \( N \) is the normality we just calculated (0.02 N)

### Step 4: Substitute the values into the formula
Substituting the values into the formula:
\[ \Lambda = \frac{1000 \times 0.000158}{0.02} \]

### Step 5: Calculate the equivalent conductance
Now, performing the calculation:
\[ \Lambda = \frac{0.158}{0.02} = 7.9 \, \Omega^{-1} \, \text{cm}^2 \, \text{equivalent}^{-1} \]

### Final Answer
The equivalent conductance of the 0.01 M alkaline earth metal chloride at 25°C is:
\[ \Lambda = 7.9 \, \Omega^{-1} \, \text{cm}^2 \, \text{equivalent}^{-1} \]

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