Question:

Number of moles of ions produced by the complete dissociation of one mole of Mohr’s salt in water is:

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To calculate the number of ions, consider the complete dissociation of all components of the salt.
Updated On: Apr 16, 2025
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The Correct Option is C

Approach Solution - 1

Mohr’s salt, chemically known as ammonium iron(II) sulfate, has the formula (NH4)2[Fe(SO4)2] · 6H2O.

When dissolved in water, it dissociates completely as follows:

\( (\text{NH}_4)_2[\text{Fe(SO}_4)_2] \cdot 6\text{H}_2\text{O} \rightarrow 2\text{NH}^+_4 + \text{Fe}^{2+} + 2\text{SO}^{2-}_4 \)

From this dissociation, the ions produced are:

  • 2 NH+4 (ammonium ions),
  • 1 Fe2+ (iron(II) ion),
  • 2 SO2−4 (sulfate ions).

Adding these, the total number of moles of ions produced from one mole of Mohr’s salt is: 2 + 1 + 2 = 5 moles of ions.

Conclusion: The correct answer is: 5.

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Approach Solution -2

Correct Answer:

Option 3: 5

Explanation:

Mohr's salt has the formula (NH₄)₂Fe(SO₄)₂·6H₂O. When it dissociates in water, it produces the following ions:

(NH₄)₂Fe(SO₄)₂·6H₂O (aq) → 2NH₄⁺(aq) + Fe²⁺(aq) + 2SO₄²⁻(aq) + 6H₂O(l)

This results in 2 ammonium ions (NH₄⁺), 1 iron(II) ion (Fe²⁺), and 2 sulfate ions (SO₄²⁻), for a total of 5 ions. The water molecules remain as water.

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