On treatment of 100 ml of 0.1 M solution of the complex CrCl 3 ⋅ 6 H 2 O with excess of AgNO 3, 4.305 g of AgCl was obtained. The complex is

Question

On treatment of 100 ml of 0.1 M solution of the complex CrCl 3 ⋅ 6 H 2 O with excess of AgNO 3, 4.305 g of AgCl was obtained. The complex is (A) [ Cr ( H 2 O )3 Cl 3] ⋅ 3 H 2 O (B) [ Cr ( H 2 O )4 Cl 2] Cl 2 H 2 O (C) [ Cr ( H 2 O )5 Cl ] Cl 2 ⋅ H 2 O (D) [ Cr ( H 2 O )6] Cl 3

Tardigrade Admin · Accepted Answer

Mol of AgCl =(4.305/143.5)=0.03= mol of Cl -given by the complex. Mol of the complex =100 × 10-3 × 0.1=0.01; underset0.01 mol[Cr(H2O)6]Cl3 arrow underset0.01 mol[Cr(H2O)6]3+ + underset0.03 mol3Cl-

Q. On treatment of $100 m l$ of $0.1 M$ solution of the complex $C r C l_{3} \cdot 6 H_{2} O$ with excess of $A g N O_{3}, 4.305 g$ of $A g Cl$ was obtained. The complex is

$[C r (H_{2} O)_{3} C l_{3}] \cdot 3 H_{2} O$

$[C r (H_{2} O)_{4} C l_{2}] C l_{2} H_{2} O$

$[C r (H_{2} O)_{5} Cl] C l_{2} \cdot H_{2} O$

$[C r (H_{2} O)_{6}] C l_{3}$

$M o l$ of $A g Cl = \frac{4.305}{143.5} = 0.03 = m o l$ of $C l^{-}$ given by the complex. Mol of the complex $= 100 \times 1 0^{- 3} \times 0.1 = 0.01$ ; $0.01 m o l [C r (H_{2} O)_{6}] C l_{3} \to 0.01 m o l [C r (H_{2} O)_{6}]^{3 +} + 0.03 m o l 3 C l^{-}$

Q. On treatment of 100ml of 0.1M solution of the complex CrCl3​⋅6H2​O with excess of AgNO3​,4.305g of AgCl was obtained. The complex is

[Cr(H2​O)3​Cl3​]⋅3H2​O

[Cr(H2​O)4​Cl2​]Cl2​H2​O

[Cr(H2​O)5​Cl]Cl2​⋅H2​O

[Cr(H2​O)6​]Cl3​