PtCl4,6H2O can exist as hydrated complex 1 molal aq. solution has depression in freezing point of 3.72∘ Assume 100% ionisation and KfH2O=1.86∘mol−1kg, then complex is

Pt H 2 O 4 Cl 2 Cl 2 ⋅ 2 H 2 O

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${PtCl}_{4}, 6 H_{2} O$ can exist as hydrated complex $1 molal aq .$ solution has depression in freezing point of ${3.72}^{\circ}$ Assume 100% $ionisation$ and $K_{f} (H_{2} O) = {1.86}^{\circ} {mol}^{- 1} kg,$ then complex is

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$[Pt {(H_{2} O)}_{4} {Cl}_{2}] {Cl}_{2} \cdot 2 H_{2} O$

$[Pt {(H_{2} O)}_{2} {Cl}_{4}] \cdot 4 H_{2} O$

$[Pt {(H_{2} O)}_{6}] {Cl}_{4}$

$[Pt {(H_{2} O)}_{3} {Cl}_{3}] Cl \cdot 3 H_{2} O$

answer is C.

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Freezing temperature is given as,

$\begin{array}{l} Δ T_{f} = i K_{f} m \end{array}$

Thus, the value of Van't Hoff factor,

$\begin{array}{l} \Rightarrow i = \frac{3.72}{1.86 \times 1} = 2 \end{array}$

This satisfy the following reaction,

$\Rightarrow [Pt {(H_{2} O)}_{3} {Cl}_{3}] Cl ⟶ [Pt {(H_{2} O_{0} {Cl}_{2}]}^{+} + {Cl}^{-}$

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