A 0.001 , text{molal} solution of [Pt(NH_3)_4 | vedclass

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(B) Given $\Delta T_f = 0.0054 \, ^\circ C$,$m = 0.001 \, 	ext{molal}$,and $K_f = 1.80$. Using the formula $\Delta T_f = i \cdot K_f \cdot m$: $0.005

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$[Pt(NH_3)_4Cl_2]Cl_2$

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(B) Given $\Delta T_f = 0.0054 \, ^\circ C$,$m = 0.001 \, 	ext{molal}$,and $K_f = 1.80$. Using the formula $\Delta T_f = i \cdot K_f \cdot m$: $0.0054 = i 	imes 1.80 	imes 0.001$ $i = \frac{0.0054}{0.0018} = 3$ $A$ van't Hoff factor of $i = 3$ indicates that the compound dissociates into $3$ ions. The formulation $[Pt(NH_3)_4Cl_2]Cl_2$ dissociates as: $[Pt(NH_3)_4Cl_2]Cl_2 ightarrow [Pt(NH_3)_4Cl_2]^{2+} + 2Cl^{-}$ This gives a total of $3$ ions ($1$ complex cation and $2$ chloride anions).

$A$ $0.001 \, \text{molal}$ solution of $[Pt(NH_3)_4Cl_4]$ in water shows a freezing point depression of $0.0054 \, ^\circ C$. If $K_f$ for water is $1.80$,the correct formulation of the above compound is?

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