JEE Main 2026 — Chemistry Inorganic Chemistry

2026integermedium

$5.33$ g of $CrCl_3 \cdot 6H_2O$ , 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 $AgNO_3$ 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 mol $^{-1}$ Cr : $52$ , Ag : $108$ , Cl : $35.5$ , H : $1$ , O : $16$ ]

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

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Solution

Molar mass of $CrCl_3 \cdot 6H_2O = 52 + 3 \times 35.5 + 6 \times 18 = 266.5 \text{ g mol}^{-1}$

Moles of complex reacted $= \dfrac{5.33}{266.5} = 0.02 \text{ mol}$

Since the complex is a $1:3$ electrolyte, it dissociates to give one complex cation and three chloride anions. Its formula is $[Cr(H_2O)_6]Cl_3$ .

When passed through a cation exchanger, the complex cation is exchanged, and the chloride ions pass into the eluted solution.

Molar mass of $AgCl = 108 + 35.5 = 143.5 \text{ g mol}^{-1}$

Moles of $AgCl$ formed $= \dfrac{8.61}{143.5} = 0.06 \text{ mol}$

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

$\text{Ratio} = \dfrac{0.02}{0.06} = \dfrac{1}{3} = 0.3333...$

This can be written as $33.33 \times 10^{-2}$ .

Rounding to the nearest integer, we get $33$ .

Answer: $33$

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