Chemistry Q71 – JEE-MAIN 2026

Home Chemistry Coordination CompoundsQuestion #325

An excess of

A g N O_{3}

is added to

100 mL

of a

0.05 M

solution of tetraaquadichloridochromium (III) chloride. The number of moles of

A g C l

precipitated will be _______

\times 10^{- 3}

. (Nearest integer)

Enter Calculated Numerical Value:

Clue 1 of 3: Coordination Chemistry

Identify the coordination complex and determine which chloride ions are outside the coordination sphere and thus available for precipitation.

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Step 1: Determine the formula of the complex and dissociable chloride ions.✦ Active

The complex "tetraaquadichloridochromium (III) chloride" has the formula $[C r (H_{2} O)_{4} C l_{2}] C l$ . When dissolved in water, it dissociates into one complex cation and one free chloride ion:

[C r (H_{2} O)_{4} C l_{2}] C l (a q) \to [C r (H_{2} O)_{4} C l_{2}]^{+} (a q) + C l^{-} (a q)

Thus, 1 mole of the complex yields 1 mole of precipitable $C l^{-}$ ions.

Step 2: Calculate the moles of the complex.○ Expand

Given volume = $100 mL = 0.100 L$ .

Given concentration = $0.05 M$ .

Moles of complex = Concentration $\times$ Volume = $0.05 mol/L \times 0.100 L = 0.005 mol$ .

Step 3: Calculate the moles of AgCl precipitated.○ Expand

Since 1 mole of the complex yields 1 mole of $C l^{-}$ ions, the moles of $C l^{-}$ ions available for precipitation are $0.005 mol$ .

The reaction with $A g N O_{3}$ is $A g^{+} (a q) + C l^{-} (a q) \to A g C l (s)$ .

Therefore, moles of $A g C l$ precipitated = moles of $C l^{-}$ ions = $0.005 mol$ .

To express this in $\times 10^{- 3}$ : $0.005 mol = 5 \times 10^{- 3} mol$ .

The nearest integer is 5.

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Chemistry Question 71 – JEE-MAIN 2026

Clue 1 of 3: Coordination Chemistry

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