The molar conductivity of a `0.5 mol//dm^(3)` solution of `AgNO_(3)` with electrolytic conductivity of `5.76 xx 10^(-3) S cm^(-1)` at `298 K` is

Options

A. (a) `2.88 S cm^(2)//mol`
B. (b) `11.52 S cm^(2)//mol`
C. (c) `0.086 S cm^(2)//mol`
D. (d) `28.8 S cm^(2)//mol`

Explanation

To find the molar conductivity of a 0.5 mol/dm³ solution of AgNO₃ with an electrolytic conductivity of 5.76 × 10⁻³ S/cm at 298 K, we can follow these steps:

### Step 1: Understand the formula for molar conductivity
The molar conductivity (Λm) can be calculated using the formula:
\[ \Lambda_m = \frac{\kappa}{C} \]
where:
- \(\Lambda_m\) = molar conductivity (S cm²/mol)
- \(\kappa\) = electrolytic conductivity (S/cm)
- \(C\) = concentration of the solution (mol/dm³)

### Step 2: Convert the concentration from mol/dm³ to mol/cm³
Given that the concentration \(C\) is 0.5 mol/dm³, we need to convert this to mol/cm³. We know that:
\[ 1 \text{ dm}^3 = 1000 \text{ cm}^3 \]
Thus,
\[ C = 0.5 \text{ mol/dm}^3 = \frac{0.5 \text{ mol}}{1000 \text{ cm}^3} = 0.5 \times 10^{-3} \text{ mol/cm}^3 \]

### Step 3: Substitute the values into the molar conductivity formula
Now we can substitute the values of \(\kappa\) and \(C\) into the molar conductivity formula:
\[ \Lambda_m = \frac{5.76 \times 10^{-3} \text{ S/cm}}{0.5 \times 10^{-3} \text{ mol/cm}^3} \]

### Step 4: Simplify the expression
When we simplify the expression, we get:
\[ \Lambda_m = \frac{5.76}{0.5} \text{ S cm}^2/\text{mol} \]
Calculating this gives:
\[ \Lambda_m = 11.52 \text{ S cm}^2/\text{mol} \]

### Step 5: Conclusion
Thus, the molar conductivity of the solution is:
\[ \Lambda_m = 11.52 \text{ S cm}^2/\text{mol} \]

### Final Answer
The answer is option B: **11.52 S cm²/mol**.

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