A complex of molecular formula CrCl3.6H2O has green colour 0.1 mole of the complex when treated with excess of AgNO3 gave 28.7 g of white precipitate of AgCl (Molar mass = 143.5 g). The formula of the complex would be

Question

A complex of molecular formula CrCl3.6H2O has green colour 0.1 mole of the complex when treated with excess of AgNO3 gave 28.7 g of white precipitate of AgCl (Molar mass = 143.5 g). The formula of the complex would be (A) [ Cr ( H 2 O )6] Cl 3 (B) [ CrCl ( H 2 O )5] Cl 2 ⋅ H 2 O (C) [ CrCl 2( H 2 O )4] Cl .2 H 2 O (D) [ Cr ( H 2 O )3 Cl 3] 3 H 2 O

Tardigrade Admin · Accepted Answer

underset0.1 m o lC r C l3 . 6 H2 O+AgNO3(excess) arrow underset28.7 gA g C l No. of moles of AgCl=(28.7/143.5)≈ 0.2mol Hence it required 2 mole free Cl atom.

Q. A complex of molecular formula $C r C l_{3} .6 H_{2} O$ has green colour $0.1$ mole of the complex when treated with excess of $A g N O_{3}$ gave $28.7 g$ of white precipitate of $A g Cl$ (Molar mass = $143.5 g$ ). The formula of the complex would be

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

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

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

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

$0.1 m o l C r C l_{3} .6 H_{2} O + A g N O_{3} (e x cess) \to 28.7 g A g Cl$

No. of moles of $A g Cl = \frac{28.7}{143.5} \approx 0.2 m o l$

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

Q. A complex of molecular formula CrCl3​.6H2​O has green colour 0.1 mole of the complex when treated with excess of AgNO3​ gave 28.7g of white precipitate of AgCl (Molar mass = 143.5g). The formula of the complex would be

[Cr(H2​O)6​]Cl3​

[CrCl(H2​O)5​]Cl2​⋅H2​O

[CrCl2​(H2​O)4​]Cl.2H2​O

[Cr(H2​O)3​Cl3​]3H2​O