Complex A has a composition of H₁₂O₆Cl₃Cr. If the complex on treatment with conc.H₂SO₄ loses 13.5% of its original mass, the correct molecular formula of A is :
[Given: atomic mass of Cr = 52 amu and Cl = 35 amu]

Solution

Chemistry ma’am · 3 steps

Conc. H₂SO₄ pulls out only the loose water outside the bracket, so find how many waters that 13.5% is.

1. Which water leaves. Conc. H₂SO₄ is a dehydrating agent. It removes only the water of crystallisation sitting outside the square bracket. Water bonded to Cr as a ligand stays put.
2. Molar mass. For H₁₂O₆Cl₃Cr: 12 + 96 + 105 + 52 = 265 g mol⁻¹, using Cl = 35 as given.

   mass lost = 13.5/100 × 265 = 35.8 ≈ 36 g = 2 × 18 g

   Option	Lattice H₂O	Mass lost	% loss
   (a)	1	18	6.8%
   (b)	2	36	13.6%
   (c)	3	54	20.4%
   (d)	0	0	0%

3. So. Exactly two water molecules must lie outside the coordination sphere, which gives [Cr(H₂O)₄Cl₂]Cl.2H₂O, option (b).

The idea to keep. Isomers with the same formula can be told apart by what leaves easily. Dehydrating agents remove only lattice water, and only ions outside the bracket precipitate with AgNO₃.

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