Question

A complex of molecular formula $CrCl_{3}.6H_{2}O$ has green colour $0.1$ mole of the complex when treated with excess of $AgNO_{3}$ gave $28.7\, g$ of white precipitate of $AgCl$ (Molar mass = $143.5\, g$). The formula of the complex would be

• A. $\left[ Cr \left( H _{2} O \right)_{6}\right] Cl _{3}$
• B. $\left[ CrCl \left( H _{2} O \right)_{5}\right] Cl _{2} \cdot H _{2} O$
• C. $\left[ CrCl _{2}\left( H _{2} O \right)_{4}\right] Cl .2 H _{2} O$
• D. $\left[ Cr \left( H _{2} O \right)_{3} Cl _{3}\right] 3 H _{2} O$

Answer 1

Answer: (B)

$\underset{0.1 m o l}{C r C l_{3} . 6 H_{2} O}+AgNO_{3}\left(excess\right) \rightarrow \underset{28.7 g}{A g C l}$

No. of moles of $AgCl=\frac{28.7}{143.5}\approx 0.2mol$

Hence it required $2$ mole free $Cl$ atom.