Question

For the complex with molecular formula $CrCl_{3}.6H_{2}O$ 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)

$\underset{0.1 \mathrm{~mol}}{\mathrm{CrCl}_3 .6 \mathrm{H}_2 \mathrm{O}}+\mathrm{AgNO}_3($ excess $) \rightarrow \underset{28.7 \mathrm{~g}}{\mathrm{AgCl}}$

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