Question

PtCl4.6H2O can exist as a hydrated complex 1 molal aq. Solution has depression in freezing point of 3.72 assume 100% ionisation and kfH20=1.86mol/kg then complex is

Answer 1

If Hexa - hydrated platinum chloride  would present as hydrated complex , 1 molal aq. solution will have depression in freezing point of 3.720. By assuming 100% ionization & Kf (H2O) = 1.860 mol-1 kg, then the complex will be given below :

[ Pt ( H2O ) 3 Cl3 ] Cl . 3 H20

Answer 2

Answer:

The complex is $[PtCl_3(H_2O)_3]Cl.3H_2O$

Explanation:

Solution has depression in freezing point ,$\Delta T_f= 3.72 K$

Molality of the solution = 1 molal

Depression in freezing point = $k_f=1.86 K mol/kg$

The formula of depression in freezing point is given by:

$\Delta T_f=i\times k_f\times m$

$3.72 K=i\times 1.86 K mol /kg\times 1 m$

i = 2

100 % ionization means that degree of dissociation is 1

$\alpha = 1$

$\alpha =\frac{i-1}{n-1}$

$1=\frac{2-1}{n-1}$

n = 2

n = number ions given by the complex on dissociation in water.

The complex gives 2 ions in water on ionization.So the complex :

$[PtCl_3(H_2O)_3]Cl.3H_2O\rightarrow [PtCl_3(H_2O)_3]^++Cl^-$