JEE Main 2026 — 6 April, Evening Shift — Question 60

$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}+\mathrm{AgNO}_{3} \rightarrow \mathrm{AgCl} \downarrow+$ $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{NO}_{3}$ $\Rightarrow 1$ mole of complex will precipitate 1 mole of AgCl . ⇒ Given moles of complex $=$ $\mathrm{M} \times \mathrm{V}=\left(0.05 \times 100 \times 10^{-3}\right)$ moles ⇒ Moles of $\mathrm{AgCl}=5 \times 10^{-3}$ moles $\Rightarrow \mathrm{x}=5$