CrCl3 xNH3 can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of 0.558 C. Assuming 100% ionisation of this complex and coordination number of Cr is 6, the complex will be (Given Kf = 1.86 K kg mol-1)
- [Cr(NH3)5Cl]Cl2
- [Cr(NH3)4Cl2]Cl
- [Cr(NH3)6]Cl3
- [Cr(NH3)3Cl3]
Answer
(1) [Cr(NH3)5Cl]Cl2
Solution
Concept: Tf = i Kf m ⇒ 0.558 = i 1.86 0.1 ⇒ i = 3 (3 ions).
[Cr(NH3)5Cl]Cl2 [Cr(NH3)5Cl]2+ + 2Cl- ⇒ 3 ions ✓
(2) gives 2 ions, (3) gives 4 ions, (4) is neutral (1).
Answer: (1)