Question

On treatment of $100\, ml$ of $0.1 \,M$ solution of the complex $CrCl _3 \cdot 6 H _2 O$ with excess of $AgNO _3, 4.305 \,g$ of $AgCl$ was obtained. The complex is

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

Answer 1

Answer: (D)

$Mol$ of $AgCl =\frac{4.305}{143.5}=0.03= mol$ of $Cl ^{-}$given by the complex. Mol of the complex $=100 \times 10^{-3} \times 0.1=0.01$; $\underset{0.01\,mol}{[Cr(H_2O)_6]Cl_3} \rightarrow\underset{0.01\,mol}{[Cr(H_2O)_6]^{3+}} + \underset{0.03\,mol}{3Cl^-}$