The elevation of boiling point of 0.10 m aqueous CrCl₃.xNH₃ solution is two times that of 0.05 m aqueous CaCl₂ solution. The value of x is ______.
[Assume 100% ionisation of the complex and CaCl₂, coordination number of Cr as 6, and that all NH₃ molecules are present inside the coordination sphere]

Solution

Chemistry ma’am · 4 steps

Match the van't Hoff factors first, then count how many chlorides must sit outside the bracket.

1. Compare the two. For the same solvent, Δ T_b = i K_b m, so Δ T_b is proportional to i × m. CaCl₂ gives 3 ions, so i = 3.

   i × 0.10 = 2 × (3 × 0.05) = 0.30 ⇒ i = 3

2. Build the complex. Cr has coordination number 6 and all x ammonias are inside, so 6 - x chlorides are also inside. The remaining 3 - (6 - x) = x - 3 chlorides are free ions.
3. Count the ions. One complex ion plus x - 3 chloride ions gives i = 1 + (x - 3) = x - 2. Setting x - 2 = 3 gives x = 5.
4. So. The complex is [Cr(NH₃)₅Cl]Cl₂, which gives 3 ions, and x = 5.

The common slip: 6. Reading i = 3 as three free chloride ions gives x = 6, but [Cr(NH₃)₆]Cl₃ actually gives 4 ions: the complex ion plus three Cl⁻.

The idea to keep. In a coordination compound, only the counter-ions outside the square bracket ionise. The complex ion counts as one particle, however many ligands it holds.

Answer from the NTA final answer key. Solution by PrepWiser.