Mohr’s Salt Conductance Trick 🔥 | Kohlrausch Law

 

❓ Question

Two statements are given:

Statement I:
Mohr's salt is composed of only three types of ions — ferrous, ammonium and sulfate.

Statement II:
If molar conductance at infinite dilution of Fe²⁺, NH₄⁺ and SO₄²⁻ are $x_1, x_2, x_3$ respectively, then molar conductance of Mohr’s salt at infinite dilution is:

$$x_1 + x_2 + 2x_3$$

Determine which statement is correct.

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🖼 Question Image

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✍️ Short Concept

Mohr’s salt is a double salt.

Formula:

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

When dissolved in water, it completely dissociates into simple ions.

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🔷 Step 1 — Identify the Ions 💯

Mohr’s salt dissociation:

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

So ions present are:

• Ferrous ion

• Ammonium ion

• Sulfate ion

Thus Statement I is correct.

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🔷 Step 2 — Kohlrausch Law

At infinite dilution:

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

Where:

$\nu_i$ = number of ions
$\lambda_i^\circ$ = ionic conductance.

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🔷 Step 3 — Apply to Mohr’s Salt

Ions and their numbers:

Fe²⁺ → 1
NH₄⁺ → 2
SO₄²⁻ → 2

So,

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

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🔷 Step 4 — Compare with Statement II

Statement II gives:

$$x_1 + x_2 + 2x_3$$

But correct expression is:

$$x_1+2x_2+2x_3$$

So Statement II is incorrect.

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✅ Final Answer

$$\boxed{\text{Statement I is correct, Statement II is incorrect}}$$

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⭐ Golden JEE Insight

For double salts:

Always write complete dissociation first.

Then apply:

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

Ion counting = answer.