JEE Main 2026 — Apr 2, Shift 1 — Chemistry solution

JEE Main2026Apr 2, Shift 1Chemistry

$5.33$ 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 $A g C l$ . The ratio of moles of complex reacted and moles of $A g C l$ formed is _____ $\times 1 0^{- 2}$ . (Nearest integer)

[Molar mass in g mol $^{- 1}$ Cr : $52$ , Ag : $108$ , Cl : $35.5$ , H : $1$ , O : $16$ ]

Your answer

Solution

As a $1 : 3$ electrolyte, the complex is $[C r (H_{2} O)_{6}] C l_{3}$ , releasing three ionisable $C l^{-}$ ions per formula unit. Each $C l^{-}$ precipitates as one $A g C l$ :

$[C r (H_{2} O)_{6}] C l_{3} + 3 A g N O_{3} \to 3 A g C l ↓ + other products$

So the stoichiometric ratio is:

$\frac{n _{complex}}{n _{A g C l}} = \frac{1}{3} = 0.333 = 33.3 \times 1 0^{- 2}$

To the nearest integer, the value is $33$ .

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