The molar ionic conductance at infinite dilution of `K^(+)` and `SO_(4)^(2-)`
are 73.5 and 160 `Scm^(2)mol^(-1)` respectively.The molar
conductance of solution of `K_(2)SO_(4)` at infinite dilution will be

Options

A. 233.5
B. 307
C. 153.5
D. 467

Explanation

To find the molar conductance of the solution of \( K_2SO_4 \) at infinite dilution, we can use the following steps:

### Step 1: Identify the ions and their molar ionic conductances
The dissociation of \( K_2SO_4 \) in water can be represented as:
\[ K_2SO_4 \rightarrow 2K^+ + SO_4^{2-} \]
From the problem, we know:
- Molar ionic conductance of \( K^+ \) = 73.5 \( \text{S cm}^2 \text{mol}^{-1} \)
- Molar ionic conductance of \( SO_4^{2-} \) = 160 \( \text{S cm}^2 \text{mol}^{-1} \)

### Step 2: Apply Kohlrausch's Law
Kohlrausch's Law states that the molar conductance of an electrolyte at infinite dilution is the sum of the molar ionic conductances of its constituent ions. Therefore, we can express the molar conductance \( \Lambda_m \) of \( K_2SO_4 \) as:
\[ \Lambda_m = \nu_+ \cdot \lambda_+ + \nu_- \cdot \lambda_- \]
Where:
- \( \nu_+ \) = number of cations = 2 (for \( K^+ \))
- \( \nu_- \) = number of anions = 1 (for \( SO_4^{2-} \))
- \( \lambda_+ \) = molar ionic conductance of \( K^+ \) = 73.5 \( \text{S cm}^2 \text{mol}^{-1} \)
- \( \lambda_- \) = molar ionic conductance of \( SO_4^{2-} \) = 160 \( \text{S cm}^2 \text{mol}^{-1} \)

### Step 3: Substitute the values into the equation
Now substituting the values into the equation:
\[ \Lambda_m = (2 \cdot 73.5) + (1 \cdot 160) \]

### Step 4: Calculate the molar conductance
Calculating the above expression:
\[ \Lambda_m = 147 + 160 = 307 \, \text{S cm}^2 \text{mol}^{-1} \]

### Final Answer
Thus, the molar conductance of the solution of \( K_2SO_4 \) at infinite dilution is:
\[ \Lambda_m = 307 \, \text{S cm}^2 \text{mol}^{-1} \]

---