Molar conductivity of a solution of an electrolyte `AB_3` is 150`Scm^2mol^(-1)` . If it ionises as `AB_(3)toA^(3+)+3B^(-)`, its equivalent conductivity will be :

Options

A. 150 (in`Scm^2eq^(-1)`)
B. 75 (in `Scm^2eq^(-1)`)
C. 50 (in `Scm^2eq^(-1)`)
D. 80 (in `Scm^2eq^(-1)`)

Answer

To solve the problem, we need to find the equivalent conductivity of the electrolyte \( AB_3 \) given its molar conductivity. Let's break it down step by step.

### Step 1: Understand the Ionization of the Electrolyte
The electrolyte \( AB_3 \) ionizes as follows:
\[ AB_3 \rightarrow A^{3+} + 3B^{-} \]
This means that one formula unit of \( AB_3 \) produces one \( A^{3+} \) ion and three \( B^{-} \) ions.

### Step 2: Determine the n-factor
The n-factor is defined as the number of moles of reactive species produced per mole of the electrolyte. In this case:
- From \( AB_3 \), we get 1 mole of \( A^{3+} \) and 3 moles of \( B^{-} \).
- Therefore, the total number of moles of ions produced is \( 1 + 3 = 4 \).

Thus, the n-factor \( n \) for \( AB_3 \) is 4.

### Step 3: Use the Relationship Between Molar Conductivity and Equivalent Conductivity
The relationship between molar conductivity (\( \Lambda_m \)) and equivalent conductivity (\( \Lambda_e \)) is given by:
\[ \Lambda_e = \frac{\Lambda_m}{n} \]
Where:
- \( \Lambda_m \) = Molar conductivity
- \( \Lambda_e \) = Equivalent conductivity
- \( n \) = n-factor

### Step 4: Substitute the Values
Given that the molar conductivity \( \Lambda_m \) is 150 S cm² mol⁻¹ and \( n \) is 4:
\[ \Lambda_e = \frac{150}{4} \]

### Step 5: Calculate the Equivalent Conductivity
Now, calculate the equivalent conductivity:
\[ \Lambda_e = \frac{150}{4} = 37.5 \, \text{S cm}^2 \text{equivalent}^{-1} \]

### Final Answer
The equivalent conductivity of the solution is:
\[ \Lambda_e = 37.5 \, \text{S cm}^2 \text{equivalent}^{-1} \]