Compound `PdCl_(4). 6H_(2)O` is a hydrated complex, 1 molal aqueous solution of it has freezing point 269.28 K. Assuming 100% ionization of complex, calculate the molecular formula of the complex (`K_(f)` for water = 1.86 K `mol^(-1)`)

Options

A. `[Pd(H_2O)_6]Cl_4`
B. `[Pd(H_2O)_4Cl_2 ]Cl_2 .2H_2O`
C. `[Pd(H_2O)_3Cl_3]Cl.3H_2O`
D. `[Pd(H_2O)_2Cl_4].4H_2O`

Answer

C

Explanation

`DeltaT=ixxK_"f"xxm`
`(273-269.28)=ixx1.86xx1`
`3.72=ixx1.86`
I = 2
`alpha=(i-1)/(n-1)=1=(2-1)/(n-1)" or "n=2`
Thus, the complex should give two ions in the solution,
i.e., the complex will be `[Pd(H_2O)_3Cl_3]Cl.3H_2O`