Chemistry Q55 – JEE-MAIN 2025

Home Chemistry ElectrochemistryQuestion #404

Given below are two statements: Statement I: Mohr's salt is composed of only three types of ions-ferrous, ammonium and sulfate. Statement II: If the molar conductance at infinite dilution for ferrous, ammonium and sulfate ions are

x_{1}

x_{2}

and

x_{3}

S cm

^{2}

mol

^{- 1}

, respectively then the molar conductance for Mohr's salt solution at infinite dilution would be given by

x_{1} + x_{2} + 2 x_{3}

In the light of the given statements, choose the correct answer from the options given below.

Clue 1 of 3: Composition of Mohr's Salt

Recall the chemical formula and dissociation products of Mohr's salt in aqueous solution.

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Step 1: Analyze Statement I: Composition of Mohr's Salt✦ Active

Mohr's salt is ammonium ferrous sulfate hexahydrate, with the chemical formula $(N H_{4})_{2} F e (S O_{4})_{2} \cdot 6 H_{2} O$ . When dissolved in water, it dissociates into $F e^{2 +}$ (ferrous ions), $N H_{4}^{+}$ (ammonium ions), and $S O_{4}^{2 -}$ (sulfate ions). Thus, it is composed of three types of ions. Therefore, Statement I is TRUE.

Step 2: Analyze Statement II: Molar Conductance using Kohlrausch's Law○ Expand

According to Kohlrausch's Law of Independent Migration of Ions, the molar conductance at infinite dilution ( $Λ_{m}^{\circ}$ ) of an electrolyte is the sum of the molar ionic conductivities of its constituent ions, each multiplied by its stoichiometric coefficient in the dissociated salt.

For Mohr's salt, $(N H_{4})_{2} F e (S O_{4})_{2}$ , the dissociation yields $1 \times F e^{2 +}$ , $2 \times N H_{4}^{+}$ , and $2 \times S O_{4}^{2 -}$ ions.

Given the molar conductances at infinite dilution for $F e^{2 +}$ , $N H_{4}^{+}$ , and $S O_{4}^{2 -}$ ions are $x_{1}$ , $x_{2}$ , and $x_{3}$ S cm $^{2}$ mol $^{- 1}$ respectively.

Λ_{m}^{\circ} (Mohr's salt) = 1 \cdot λ_{F e^{2 +}}^{\circ} + 2 \cdot λ_{N H_{4}^{+}}^{\circ} + 2 \cdot λ_{S O_{4}^{2 -}}^{\circ}

Λ_{m}^{\circ} (Mohr's salt) = x_{1} + 2 x_{2} + 2 x_{3}

Statement II claims the molar conductance would be $x_{1} + x_{2} + 2 x_{3}$ . This is incorrect because there are two ammonium ions, so the contribution from ammonium ions should be $2 x_{2}$ , not $x_{2}$ . Therefore, Statement II is FALSE.

💡 Teacher's Secret Hint

Pay close attention to the stoichiometric coefficients of each ion when applying Kohlrausch's Law.

Step 3: Determine the Correct Option○ Expand

Based on the analysis, Statement I is True and Statement II is False. This corresponds to option 3.

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Chemistry Question 55 – JEE-MAIN 2025

Clue 1 of 3: Composition of Mohr's Salt

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