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. $\left[ Cr \left( H _{2} O \right)_{4} Cl _{2}\right] Cl \cdot 2 H _{2} O$
• 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 $=100 \times 10^{-3} \times 0.1=0.01$

$\underset{0.01 \,mol}{[ Cr ( H _{2} O )_{6}] }Cl _{3} \longrightarrow \underset{0.01 \,mol}{[ Cr ( H _{2} O )_{6}]^{3+}}+\underset{0.03 \,mol }{3 Cl ^{-}}$