0.0012 mol of CrCl 3 ⋅ 6 H 2 O was passed through a cation exchange resin and acid coming out of it required 28.5 mL of 0.125 M NaOH. Hence, complex is

Question

0.0012 mol of CrCl 3 ⋅ 6 H 2 O was passed through a cation exchange resin and acid coming out of it required 28.5 mL of 0.125 M NaOH. Hence, complex is (A) [ Cr ( H 2 O )5 Cl 2 ⋅ H 2 O . (B) [ Cr ( H 2 O )4 Cl 2] ⋅ 2 H 2 O (C) [ Cr ( H 2 O )6] Cl 3 (D) [ Cr ( H 2 O )3 Cl 3] ⋅ 3 H 2 O

Tardigrade Admin · Accepted Answer

Correct answer is (c) [ Cr ( H 2 O )6] Cl 3

Q. $0.0012 m o l$ of $C r C l_{3} \cdot 6 H_{2} O$ was passed through a cation exchange resin and acid coming out of it required $28.5 m L$ of $0.125 M N a O H$ . Hence, complex is

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

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

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

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

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

Q. 0.0012mol of CrCl3​⋅6H2​O was passed through a cation exchange resin and acid coming out of it required 28.5mL of 0.125MNaOH. Hence, complex is

[Cr(H2​O)5​Cl2​⋅H2​O

[Cr(H2​O)4​Cl2​]⋅2H2​O

[Cr(H2​O)6​]Cl3​

[Cr(H2​O)3​Cl3​]⋅3H2​O