For the complex with molecular formula CrCl3.6H2O 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

For the complex with molecular formula CrCl3.6H2O 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 mathrm~mol mathrmCrCl3 .6 mathrmH2 mathrmO+ mathrmAgNO3( excess ) arrow underset28.7 mathrm~g mathrmAgCl No. of moles of AgCl=(28.7/143.5)≈ 0.2mol

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

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

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

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

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

$0.1 mol CrCl_{3} .6 H_{2} O + AgNO_{3} ($ excess $) \to 28.7 g AgCl$

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

Q. For the complex with molecular formula CrCl3​.6H2​O 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

[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