JEE Chemistry 2026 Question with Solution

Given: Volume of potassium dichromate solution = $200 \text{ cc}$, molarity = $x \times 10^{-3} \text{ M}$. Volume of Mohr's salt solution = $750 \text{ cc}$, molarity = $0.6 \text{ M}$.

Find: The value of $x$.

Use the normality relation:

$$N_1 V_1 = N_2 V_2$$

For $\text{K}_2\text{Cr}_2\text{O}_7$, n-factor $= 6$, so

$$N_1 = 6 \times M_1 = 6x \times 10^{-3}$$

For Mohr's salt, $\text{Fe}^{2+} \to \text{Fe}^{3+}$ and n-factor $= 1$, so

$$N_2 = 1 \times M_2 = 0.6$$

Substitute the given volumes:

$$(6x \times 10^{-3}) \times 200 = 0.6 \times 750$$ $$1200x \times 10^{-3} = 450$$ $$1.2x = 450$$ $$x = \frac{450}{1.2} = 375$$

Therefore, the value of $x$ is $375$.

Given: Dichromate acts as oxidizing agent in acidic medium and $\text{Fe}^{2+}$ from Mohr's salt is oxidized to $\text{Fe}^{3+}$.

Find: The value of $x$ in the molarity of potassium dichromate.

The relevant n-factors are:

• $\text{Cr}_2\text{O}_7^{2-} \to \text{Cr}^{3+}$ gives n-factor $= 6$
• $\text{Fe}^{2+} \to \text{Fe}^{3+}$ gives n-factor $= 1$

Hence, normality of potassium dichromate solution is

$$N_1 = 6 \times x \times 10^{-3}$$

and normality of Mohr's salt solution is

$$N_2 = 0.6$$

Now apply

$$N_1V_1 = N_2V_2$$

with $V_1 = 200$ and $V_2 = 750$:

$$(6x \times 10^{-3})(200) = (0.6)(750)$$ $$1200x \times 10^{-3} = 450$$ $$1.2x = 450$$ $$x = 375$$

Thus, the required value is $375$.