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(B) The formula for freezing point depression is $\Delta T_f = i 	imes K_f 	imes m$. Given: $\Delta T_f = 0.0054$,$K_f = 1.80$,$m = 0.001$. Substitu

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$[Pt(NH_3)_4 Cl_2] Cl_2$

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(B) The formula for freezing point depression is $\Delta T_f = i 	imes K_f 	imes m$. Given: $\Delta T_f = 0.0054$,$K_f = 1.80$,$m = 0.001$. Substituting the values: $0.0054 = i 	imes 1.80 	imes 0.001$. $i = \frac{0.0054}{0.0018} = 3$. Since the van't Hoff factor $i = 3$,the complex dissociates into $3$ ions. The complex $[Pt(NH_3)_4 Cl_2] Cl_2$ dissociates as $[Pt(NH_3)_4 Cl_2]^{2+} + 2Cl^-$,providing $3$ ions. Therefore,the correct formula is $[Pt(NH_3)_4 Cl_2] Cl_2$.

The freezing point depression of a $0.001 \ m$ aqueous solution of $[Pt(NH_3)_4 Cl_4]$ is $0.0054^{\circ} C$. If $K_f$ for water is $1.80 \ K \ kg \ mol^{-1}$,determine the correct formula of the complex.

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