Given the ionic equivalent conductivities for the following ions `:`
`lambda^(@)._(eq)K^(o+)=73.5cm^(2)ohm^(-1)eq^(-1)`
`lambda^(@)._(eq)Al^(3+)=149cm^(2)ohm^(-1)eq^(-1)`
`lambda^(@)._(eq)SO_(4)^(2-)=85.8cm^(2)ohm^(-1)eq^(-1)`
The `wedge_(eq)^(@)` for potash alum `(K_(2)SO_(4).Al_(2)(SO_(4))_(3).24H_(2)O)` is `:`

Options

A. `215.92`
B. `348.3`
C. `368.2`
D. `108.52`

Explanation

To find the equivalent conductivity (λ^(@)_eq) for potash alum (K₂SO₄·Al₂(SO₄)₃·24H₂O), we will follow these steps:

### Step 1: Determine the total charge of the ions in potash alum
Potash alum consists of:
- 2 K⁺ ions
- 2 Al³⁺ ions
- 4 SO₄²⁻ ions

The total positive charge contributed by the ions is:
- Charge from K⁺: 2 × 1 = 2
- Charge from Al³⁺: 2 × 3 = 6
- Total positive charge = 2 + 6 = 8

The total negative charge contributed by the ions is:
- Charge from SO₄²⁻: 4 × 2 = 8
- Total negative charge = 8

### Step 2: Calculate the equivalence of each ion
- For K⁺:
\[ \text{Equivalence of K} = \frac{1 \text{ (charge of K)}}{8 \text{ (total charge)}} \times 2 \text{ (number of K ions)} = \frac{2}{8} = \frac{1}{4} \text{ equivalent} \]

- For Al³⁺:
\[ \text{Equivalence of Al} = \frac{3 \text{ (charge of Al)}}{8 \text{ (total charge)}} \times 2 \text{ (number of Al ions)} = \frac{6}{8} = \frac{3}{4} \text{ equivalent} \]

- For SO₄²⁻:
\[ \text{Equivalence of SO₄} = \frac{2 \text{ (charge of SO₄)}}{8 \text{ (total charge)}} \times 4 \text{ (number of SO₄ ions)} = \frac{8}{8} = 1 \text{ equivalent} \]

### Step 3: Calculate the equivalent conductivity (λ^(@)_eq) for potash alum
Using the formula:
\[ \lambda^(@)_eq = \left( \text{Equivalence of K} \times \lambda^(@)_eq K^+ \right) + \left( \text{Equivalence of Al} \times \lambda^(@)_eq Al^{3+} \right) + \left( \text{Equivalence of SO₄} \times \lambda^(@)_eq SO₄^{2-} \right) \]

Substituting the values:
- λ^(@)_eq K⁺ = 73.5 cm²Ω⁻¹eq⁻¹
- λ^(@)_eq Al³⁺ = 149 cm²Ω⁻¹eq⁻¹
- λ^(@)_eq SO₄²⁻ = 85.8 cm²Ω⁻¹eq⁻¹

Calculating each term:
\[ \lambda^(@)_eq = \left( \frac{1}{4} \times 73.5 \right) + \left( \frac{3}{4} \times 149 \right) + \left( 1 \times 85.8 \right) \]

Calculating:
\[ = 18.375 + 111.75 + 85.8 \]
\[ = 215.925 \text{ cm²Ω⁻¹eq⁻¹} \]

### Final Answer
The equivalent conductivity (λ^(@)_eq) for potash alum is approximately **215.93 cm²Ω⁻¹eq⁻¹**.

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