Q21 / 75 · JEE Main 2026 (06 April Shift 2) · Chemistry
An excess of AgNO$_3$ is added to $100$ mL of a $0.05$ M solution of tetraaquadichloridochromium (III) chloride. The number of moles of AgCl precipitated will be __________ $\times 10^{-3}$. (Nearest integer)
Correct answer: 5
Solution
The formula of tetraaquadichloridochromium (III) chloride is $[Cr(H_2O)_4Cl_2]Cl$.
In this complex, there is one ionizable chloride ion present outside the coordination sphere.
The dissociation reaction is:
$[Cr(H_2O)_4Cl_2]Cl \rightarrow [Cr(H_2O)_4Cl_2]^+ + Cl^-$
Number of moles of the complex $= M \times V = 0.05 \times \dfrac{100}{1000} = 0.005$ moles.
Since one mole of the complex yields one mole of chloride ions, the number of moles of $Cl^-$ ions produced is $0.005$ moles.
When excess $AgNO_3$ is added, the chloride ions precipitate as $AgCl$:
$Ag^+ + Cl^- \rightarrow AgCl \downarrow$
Number of moles of $AgCl$ precipitated $= 0.005 = 5 \times 10^{-3}$ moles.
The value is $5$.
Answer: $5$
In this complex, there is one ionizable chloride ion present outside the coordination sphere.
The dissociation reaction is:
$[Cr(H_2O)_4Cl_2]Cl \rightarrow [Cr(H_2O)_4Cl_2]^+ + Cl^-$
Number of moles of the complex $= M \times V = 0.05 \times \dfrac{100}{1000} = 0.005$ moles.
Since one mole of the complex yields one mole of chloride ions, the number of moles of $Cl^-$ ions produced is $0.005$ moles.
When excess $AgNO_3$ is added, the chloride ions precipitate as $AgCl$:
$Ag^+ + Cl^- \rightarrow AgCl \downarrow$
Number of moles of $AgCl$ precipitated $= 0.005 = 5 \times 10^{-3}$ moles.
The value is $5$.
Answer: $5$
Hence, the correct answer is 5.