What will be the molar conductivity of Al 3+ ions at infinite dilution if molar conductivity of `Al^(2) (SO_(4))_(3)` is 858 S `cm^(2)` `"mol" ^(-1)` and ionic conductance of `SO_(4)^(2-)` is 160 S `cm^(2) "mol" ^(-1)` at infinite dilution ?

Options

A. `189 S cm^(2) "mol"^(-1)`
B. `698 S cm^(2) "mol"^(-1)`
C. `1018 S cm^(2) "mol"^(-1)`
D. `429 S cm^(2) "mol"^(-1)`

Explanation

To find the molar conductivity of Al³⁺ ions at infinite dilution, we will use Kohlrausch's law of independent migration of ions. Here’s a step-by-step solution:

### Step 1: Understand the given data
We have the following information:
- Molar conductivity of Al₂(SO₄)₃ at infinite dilution, \( \Lambda_m(Al_2(SO_4)_3) = 858 \, S \, cm^2 \, mol^{-1} \)
- Ionic conductance of SO₄²⁻, \( \lambda_0(SO_4^{2-}) = 160 \, S \, cm^2 \, mol^{-1} \)

### Step 2: Write the formula using Kohlrausch's law
According to Kohlrausch's law, the molar conductivity of the salt can be expressed as the sum of the molar conductivities of its constituent ions:
\[ \Lambda_m(Al_2(SO_4)_3) = 2\lambda_0(Al^{3+}) + 3\lambda_0(SO_4^{2-}) \]

### Step 3: Substitute the known values into the equation
Let \( \lambda_0(Al^{3+}) = x \). Then we can write the equation as:
\[ 858 = 2x + 3 \times 160 \]

### Step 4: Calculate the contribution of SO₄²⁻
Calculate \( 3 \times 160 \):
\[ 3 \times 160 = 480 \]

### Step 5: Substitute back into the equation
Now substitute this value back into the equation:
\[ 858 = 2x + 480 \]

### Step 6: Solve for \( x \)
Rearranging the equation gives:
\[ 2x = 858 - 480 \]
\[ 2x = 378 \]
\[ x = \frac{378}{2} = 189 \]

### Step 7: Conclusion
Thus, the molar conductivity of Al³⁺ ions at infinite dilution is:
\[ \lambda_0(Al^{3+}) = 189 \, S \, cm^2 \, mol^{-1} \]