5.33 g of $\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$, 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 $\mathrm{AgNO}_{3}$ 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 $\mathrm{g} \mathrm{mol}^{-1}$ Cr: 52, Ag: 108, Cl: 35.5, H: 1, O: 16]

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Coordination Compounds

5.33 g of

CrCl_{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

AgNO_{3}

results in 8.61 g of AgCl. The ratio of moles of complex reacted and moles of AgCl 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]

Answer: 33

The salt behaves as a

1 : 3

electrolyte,

[Cr (H_{2} O)_{6}] C l_{3}

, so each formula unit releases three chloride ions that are precipitated as AgCl. The molar mass of

CrC l_{3} \cdot 6 H_{2} O

52 + 3 (35.5) + 6 (18) = 266.5 g mo l^{- 1}

, giving moles of complex

= \frac{5.33}{266.5} = 0.02

. The molar mass of AgCl is

143.5 g mo l^{- 1}

, giving moles of AgCl

= \frac{8.61}{143.5} = 0.06

. The ratio of moles of complex to moles of AgCl is

\frac{0.02}{0.06} = \frac{1}{3} = 33.33 \times 1 0^{- 2}

, which to the nearest integer is

33

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Integer answer