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. `2.88 S cm^(2)//mol`
B. `11.52 S cm^(2)//mol`
C. `0.086 S cm^(2)//mol`
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 x 10^(-3) S/cm at 298 K, we can follow these steps:

### Step 1: Understand the relationship between conductivity and molar conductivity
The molar conductivity (Λ) is defined as the conductivity (κ) divided by the concentration (C) of the solution. The formula is:

\[ \Lambda = \frac{\kappa}{C} \]

### Step 2: Convert the concentration from mol/dm³ to mol/cm³
Given that the concentration is 0.5 mol/dm³, we need to convert this to mol/cm³. Since 1 dm³ = 1000 cm³, we can convert the concentration as follows:

\[ C = 0.5 \, \text{mol/dm}^3 = \frac{0.5 \, \text{mol}}{1000 \, \text{cm}^3} = 0.5 \times 10^{-3} \, \text{mol/cm}^3 \]

### Step 3: Use the given conductivity value
The conductivity (κ) is given as:

\[ \kappa = 5.76 \times 10^{-3} \, \text{S/cm} \]

### Step 4: Substitute the values into the molar conductivity formula
Now we can substitute the values of κ and C into the molar conductivity formula:

\[ \Lambda = \frac{5.76 \times 10^{-3} \, \text{S/cm}}{0.5 \times 10^{-3} \, \text{mol/cm}^3} \]

### Step 5: Calculate the molar conductivity
Perform the division:

\[ \Lambda = \frac{5.76 \times 10^{-3}}{0.5 \times 10^{-3}} = \frac{5.76}{0.5} = 11.52 \, \text{S cm}^2/\text{mol} \]

### Final Answer
The molar conductivity of the 0.5 mol/dm³ solution of AgNO₃ is:

\[ \Lambda = 11.52 \, \text{S cm}^2/\text{mol} \]

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