Molar ionic conductivites of a bivalent electrolyte are 57 and 73. the molar conductivity of the solution will be

Options

A. `130S cm^(2) "mol"^(-1)`
B. `65.S cm^(2)"mol"^(-1)`
C. `260S cm^(2)"mol"^(-1)`
D. `187S cm^(2)"mol"^(-1)`

Explanation

To solve the problem of finding the molar conductivity of a bivalent electrolyte given the molar ionic conductivities, follow these steps:

### Step 1: Understand the Concept
Molar conductivity (\( \Lambda_m \)) of an electrolyte is the sum of the molar ionic conductivities of its constituent ions. For a bivalent electrolyte, it can be represented as:
\[ \Lambda_m = \lambda_{cation} + \lambda_{anion} \]
where \( \lambda_{cation} \) and \( \lambda_{anion} \) are the molar ionic conductivities of the cation and anion, respectively.

### Step 2: Identify Given Values
From the problem, we are given:
- Molar ionic conductivity of the cation (\( \lambda_{cation} \)) = 57 S cm²/mol
- Molar ionic conductivity of the anion (\( \lambda_{anion} \)) = 73 S cm²/mol

### Step 3: Calculate Molar Conductivity
Now, substitute the values into the formula for molar conductivity:
\[ \Lambda_m = 57 + 73 \]
Calculating this gives:
\[ \Lambda_m = 130 \, \text{S cm}^2/\text{mol} \]

### Step 4: State the Final Answer
The molar conductivity of the solution is:
\[ \Lambda_m = 130 \, \text{S cm}^2/\text{mol} \]