A sample of 2.665 g of green $CrCl X_{3} \cdot 6 H X_{2} O$ (molar mass 266.5 g/mol, Cr coordination number 6) gives 1.435 g of $AgCl$ (molar mass 143.5 g/mol) with excess $AgNO X_{3}$ . The mass of lattice (non-coordinated) water per sample, in g, is ( $H X_{2} O = 18$ g/mol)

Answer: A

Moles of complex

= 2.665/266.5 = 0.0100

and moles of

AgCl = 1.435/143.5 = 0.0100

, so 1 free chloride per formula unit. The complex is

[CrCl X_{2} (H X_{2} O) X_{4}] Cl \cdot 2 H X_{2} O

with 2 lattice waters. Mass

= 0.0100 \times 2 \times 18 = 0.36