Chemistry Q71 – JEE-MAIN 2026

Home Chemistry Coordination CompoundsQuestion #319

5.33 g of

C r C l_{3} \cdot 6 H_{2} O

, which is a 1 : 3 electrolyte, is dissolved in water and is passed through a cation exchanger. The chloride ions in the eluted solution, on treatment with

A g N O_{3}

results in 8.61 g of AgCl. The ratio of moles of complex reacted and moles of AgCl formed is _______

\times 10^{- 2}

. (Nearest integer) [Molar mass in g

m o l^{- 1}

Cr : 52, Ag : 108, Cl : 35.5, H : 1, O : 16]

Enter Calculated Numerical Value:

Clue 1 of 3: Complex Structure and Electrolyte Type

Determine the coordination complex structure based on the given formula and its classification as a 1:3 electrolyte. A 1:3 electrolyte means it dissociates into one cation and three anions.

🧠 Full Solution Path

Step 1: Determine the structure of the complex and calculate its moles✦ Active

The complex is $C r C l_{3} \cdot 6 H_{2} O$ . Since it is a 1:3 electrolyte, it dissociates into one complex cation and three chloride anions. Therefore, its structure is $[C r (H_{2} O)_{6}] C l_{3}$ .

Molar mass of $C r C l_{3} \cdot 6 H_{2} O$ :

M_{c o m p l e x} = C r + 3 \times C l + 6 \times (2 \times H + O)

M_{c o m p l e x} = 52 + 3 \times 35.5 + 6 \times (2 \times 1 + 16)

M_{c o m p l e x} = 52 + 106.5 + 6 \times 18 = 52 + 106.5 + 108 = 266.5 g/mol

Moles of complex reacted:

n_{c o m p l e x} = \frac{Mass of complex}{M_{c o m p l e x}} = \frac{5.33 g}{266.5 g/mol} = 0.02 mol

Step 2: Calculate moles of AgCl formed○ Expand

Molar mass of AgCl:

M_{A g C l} = A g + C l = 108 + 35.5 = 143.5 g/mol

Moles of AgCl formed:

n_{A g C l} = \frac{Mass of AgCl}{M_{A g C l}} = \frac{8.61 g}{143.5 g/mol} = 0.06 mol

💡 Teacher's Secret Hint

Remember that the chloride ions in the eluted solution are the free chloride ions from the complex, which react with $A g N O_{3}$ to form AgCl.

Step 3: Calculate the required ratio and express it in the given format○ Expand

From the complex structure $[C r (H_{2} O)_{6}] C l_{3}$ , 1 mole of complex yields 3 moles of free $C l^{-}$ . The reaction with $A g N O_{3}$ is $C l^{-} (a q) + A g N O_{3} (a q) \to A g C l (s) + N O_{3}^{-} (a q)$ . Thus, moles of free $C l^{-}$ equals moles of AgCl formed.

The ratio of moles of complex reacted to moles of AgCl formed is:

Ratio = \frac{n_{c o m p l e x}}{n_{A g C l}} = \frac{0.02 mol}{0.06 mol} = \frac{1}{3} \approx 0.3333

The question asks for this ratio in the format $X \times 10^{- 2}$ . So, we set:

X \times 10^{- 2} = 0.3333

Solving for $X$ :

X = 0.3333 \times 100 = 33.33

Rounding to the nearest integer, $X = 33$ .

💡 Teacher's Secret Hint

Ensure you correctly interpret the '1:3 electrolyte' information to determine the number of free chloride ions.

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Chemistry Question 71 – JEE-MAIN 2026

Clue 1 of 3: Complex Structure and Electrolyte Type

🧠 Full Solution Path