PtCl_4 cdot 6H_2O can exist as a hydrated com | vedclass

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Question

(C) The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$. Given $\Delta T_f = 3.72$,$m = 1 \ m$,and $K_f = 1.86 \ ^\ci

Vedclass · Accepted Answer

$[Pt(H_2O)_3Cl_3]Cl \cdot 3H_2O$

Vedclass · Answer

(C) The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$. Given $\Delta T_f = 3.72$,$m = 1 \ m$,and $K_f = 1.86 \ ^\circ C \ kg \ mol^{-1}$. Substituting these values: $3.72 = i 	imes 1.86 	imes 1$. Thus,the van't Hoff factor $i = 2$. An $i$ value of $2$ indicates that the complex dissociates into $2$ ions in solution. For the complex $[Pt(H_2O)_4Cl_2]Cl_2 \cdot 2H_2O$,the dissociation is $[Pt(H_2O)_4Cl_2]Cl_2 \cdot 2H_2O ightarrow [Pt(H_2O)_4Cl_2]^{2+} + 2Cl^-$,which gives $3$ ions $(i=3)$. For the complex $[Pt(H_2O)_3Cl_3]Cl \cdot 3H_2O$,the dissociation is $[Pt(H_2O)_3Cl_3]Cl \cdot 3H_2O ightarrow [Pt(H_2O)_3Cl_3]^+ + Cl^-$,which gives $2$ ions $(i=2)$. Therefore,the correct complex is $[Pt(H_2O)_3Cl_3]Cl \cdot 3H_2O$.

$PtCl_4 \cdot 6H_2O$ can exist as a hydrated complex. Its $1 \ m$ aqueous solution has a depression in freezing point of $3.72 \ ^\circ C$. Assuming $100\%$ ionization and $K_f \ (H_2O) = 1.86 \ ^\circ C \ kg \ mol^{-1}$,the correct formula for the complex is:

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