Question

A solution containing 2.665 g of CrCl₃ .6H₂O is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO₃ to give 2.87 g of AgCl. The structure of compound is

(a) [CrCI.(H₂O)₅]CI₂ .H₂O

(b) [CrCI₂ .(H₂O)₄] CI.H₂O

(c) [CrCI.(H₂O₅)]CI.H₂O

(d) [CrCI₂ .(H₂O)₅]CI .H₂O

Answer 1

Answer is (a) [CrCI.(H₂O)₅]CI₂ .H₂O

Number of moles of CI^- ions ionised from one mole of CrCI₃ .6H₂O.

= \(\frac{2.665}{2665}=0.01\) [∵ Molecular mass of CrCI₃ .H₂O = 266.5]

∴ Moles of AgCl obtained

= Moles of Cl^- ionised

= \(\frac{2.87}{143.5}\)

= 0.02

∴ 0.01 mole of complex CrCI₃ .6H₂O gives 0.02 mole of \(C\overline I\) on ionisation.

Thus, the fomula of the complex is

[CrCI (H₂O)5] CI₂ .H₂O.