JEE Main 2026 — Chemistry Physical Chemistry

2026integermedium

200 cc of $x \times 10^{-3} \mathrm{M}$ potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's salt solution in acidic medium. Here $x=$ $\_\_\_\_$ .

Official previous-year question

Held on 23 Jan 2026 · Verified 6 Jul 2026.

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Solution

Mohr's salt ( $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$ ) is a source of $Fe^{2+}$ ions.

The redox reaction with potassium dichromate in acidic medium is:

$6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ \to 6Fe^{3+} + 2Cr^{3+} + 7H_2O$ .

Moles of $Fe^{2+}$ from Mohr's salt solution:

$n(Fe^{2+}) = 0.6 \text{ M} \times 0.750 \text{ L} = 0.45 \text{ mol}$ .

From the stoichiometry, 6 moles of $Fe^{2+}$ require 1 mole of $Cr_2O_7^{2-}$ .

Moles of $Cr_2O_7^{2-}$ required:

$n(Cr_2O_7^{2-}) = \frac{0.45}{6} = 0.075 \text{ mol}$ .

Molarity of potassium dichromate:

$M = \frac{0.075 \text{ mol}}{0.200 \text{ L}} = 0.375 \text{ M} = 375 \times 10^{-3} \text{ M}$ .

Therefore, $x = 375$ .

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Solution

Mohr's salt ( $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$ ) is a source of $Fe^{2+}$ ions.

The redox reaction with potassium dichromate in acidic medium is:

$6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ \to 6Fe^{3+} + 2Cr^{3+} + 7H_2O$ .

Moles of $Fe^{2+}$ from Mohr's salt solution:

$n(Fe^{2+}) = 0.6 \text{ M} \times 0.750 \text{ L} = 0.45 \text{ mol}$ .

From the stoichiometry, 6 moles of $Fe^{2+}$ require 1 mole of $Cr_2O_7^{2-}$ .

Moles of $Cr_2O_7^{2-}$ required:

$n(Cr_2O_7^{2-}) = \frac{0.45}{6} = 0.075 \text{ mol}$ .

Molarity of potassium dichromate:

$M = \frac{0.075 \text{ mol}}{0.200 \text{ L}} = 0.375 \text{ M} = 375 \times 10^{-3} \text{ M}$ .

Therefore, $x = 375$ .

JEE Main 2026 — Chemistry Physical Chemistry

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JEE Main 2026 — Chemistry Physical Chemistry

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Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

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