A 0.001 molal solution of [Pt(NH3)4cl4] in water had a freezing point depression

Question

A 0.001 molal solution of

[P t (N H_{3})_{4} c l_{4}]

in water had a freezing point depression of

{0.0054}^{\circ}

. If

K_{f}

for water is 1.80, the correct formulation for the above molecule is

Options:

A . [Pt(NH3)4Cl3]Cl

B . [Pt(NH3)4]Cl4

C . [Pt(NH3)4Cl2]Cl3

D . [Pt(NH3)4Cl2]Cl2

Answer: Option D
:
D

△ T = i K_{f} m

\Rightarrow i = \frac{△ T_{f}}{K_{f} m} = \frac{0.0054}{1.80 \times 0.001} = 3

Thus complexproducesthreeionspermolecule.hence possiblecorrectformulais

[P t (N H_{3})_{4} C l_{2}] C l_{2}

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