Molar Conductance of Double Salt

 

❓ Concept Question

How do we calculate molar conductance at infinite dilution for a double salt like Mohr’s salt?

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🖼 Concept Image

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✍️ Short Concept

Double salts completely dissociate into simple ions in water.

👉 Each ion contributes independently to conductance.

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🔷 Step 1 — Mohr’s Salt Basics 💯

Formula:

$$(NH_4)_2Fe(SO_4)_2 \cdot 6H_2O$$

👉 It is a double salt

So in solution:

$$\text{Complete dissociation}$$

👉 No complex formation.

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🔷 Step 2 — Types of Ions Present

Dissociation:

$$(NH_4)_2Fe(SO_4)_2 \rightarrow Fe^{2+} + 2NH_4^+ + 2SO_4^{2-}$$

Ions present:

• Fe²⁺ → 1
• NH₄⁺ → 2
• SO₄²⁻ → 2

⚠️ Always count number of ions correctly.

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🔷 Step 3 — Molar Conductance Rule

At infinite dilution:

$$\Lambda_m^\circ = \sum \nu_i \lambda_i^\circ$$

👉 Multiply each ionic conductance by its stoichiometric coefficient.

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🔷 Step 4 — Apply to Mohr’s Salt

Let:

$$\lambda^\circ(Fe^{2+}) = x_1$$
$$\lambda^\circ(NH_4^+) = x_2$$
$$\lambda^\circ(SO_4^{2-}) = x_3$$

Then:

$$\Lambda_m^\circ = x_1 + 2x_2 + 2x_3$$

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🔷 Step 5 — JEE Golden Trap

❌ Only ion types count karna

❌ Number ignore kar dena

Correct:

👉 Multiplicity matters

👉 Each ion’s contribution = coefficient × conductance

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✅ Final Takeaway

$$\boxed{\Lambda_m^\circ = x_1 + 2x_2 + 2x_3}$$

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⭐ Golden JEE Insight

Double salt:

👉 Fully dissociates

👉 Behaves like simple electrolyte mixture

$$\text{Double salt} \neq \text{Complex compound}$$