Question

$0.0012\, mol$ of $CrCl _3 \cdot 6 H _2 O$ was passed through a cation exchange resin and acid coming out of it required $28.5\, mL$ of $0.125\, M \, NaOH$. Hence, complex is

• A. $\left[ Cr \left( H _2 O \right)_5 Cl _2 \cdot H _2 O \right.$
• B. $\left[ Cr \left( H _2 O \right)_4 Cl _2\right] \cdot 2 H _2 O$
• C. $\left[ Cr \left( H _2 O \right)_6\right] Cl _3$
• D. $\left[ Cr \left( H _2 O \right)_3 Cl _3\right] \cdot 3 H _2 O$

Answer 1

Answer: (C)

Correct answer is (c) $\left[ Cr \left( H _2 O \right)_6\right] Cl _3$