The molar conducitviy of a `0.1 mol//dm^(3)` solution of `KNO_(3)` with electrolytic conductivity of `4xx10^(-3)Scm^(-1)` at 298 K is :-

Options

A. `11.52 Scm^(2)mol^(-1)`
B. `20 Scm^(2)mol^(-1)`
C. `40 Scm^(2)mol^(-1)`
D. `13.48 Scm^(2)mol^(-1)`

Answer

To find the molar conductivity of a \(0.1 \, \text{mol/dm}^3\) solution of \(KNO_3\) with an electrolytic conductivity of \(4 \times 10^{-3} \, \text{S/cm}\) at 298 K, we can follow these steps:

### Step 1: Understand the formula for molar conductivity
The molar conductivity (\(\Lambda_m\)) is calculated using the formula:
\[ \Lambda_m = \frac{\kappa}{C} \]
where:
- \(\kappa\) is the electrolytic conductivity (in S/cm),
- \(C\) is the molarity of the solution (in mol/dm³).

### Step 2: Convert the molarity from dm³ to cm³
Since \(1 \, \text{dm}^3 = 1000 \, \text{cm}^3\), we need to convert the molarity from mol/dm³ to mol/cm³:
\[ C = 0.1 \, \text{mol/dm}^3 = \frac{0.1}{1000} \, \text{mol/cm}^3 = 0.1 \times 10^{-3} \, \text{mol/cm}^3 \]

### Step 3: Substitute the values into the formula
Now we can substitute the values into the formula for molar conductivity:
\[ \Lambda_m = \frac{4 \times 10^{-3} \, \text{S/cm}}{0.1 \times 10^{-3} \, \text{mol/cm}^3} \]

### Step 4: Perform the calculation
Calculating the above expression:
\[ \Lambda_m = \frac{4 \times 10^{-3}}{0.1 \times 10^{-3}} = \frac{4}{0.1} = 40 \, \text{S cm}^2/\text{mol} \]

### Final Answer
Thus, the molar conductivity of the \(0.1 \, \text{mol/dm}^3\) solution of \(KNO_3\) is:
\[ \Lambda_m = 40 \, \text{S cm}^2/\text{mol} \]