Compound PdCl_4 cdot 6H_2O is a hydrated comp | vedclass

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Question

(C) The depression in freezing point is given by $\Delta T_f = T_f^{\circ} - T_f = 273 \ K - 269.28 \ K = 3.72 \ K$. Using the formula $\Delta T_f = i

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

Vedclass · Answer

(C) The depression in freezing point is given by $\Delta T_f = T_f^{\circ} - T_f = 273 \ K - 269.28 \ K = 3.72 \ K$. Using the formula $\Delta T_f = i 	imes K_f 	imes m$,where $m = 1 \ m$ and $K_f = 1.86 \ K \ kg \ mol^{-1}$: $3.72 = i 	imes 1.86 	imes 1 \implies i = \frac{3.72}{1.86} = 2$. Since the complex undergoes $100\%$ ionization,the van't Hoff factor $i$ represents the total number of ions produced per formula unit. For $i = 2$,the complex must dissociate into $2$ ions. Option $B$ is $[Pd(H_2O)_4Cl_2]Cl_2 \cdot 2H_2O$,which dissociates as $[Pd(H_2O)_4Cl_2]Cl_2 \cdot 2H_2O ightarrow [Pd(H_2O)_4Cl_2]^{2+} + 2Cl^-$. This gives $3$ ions $(i=3)$. Option $C$ is $[Pd(H_2O)_3Cl_3]Cl \cdot 3H_2O$,which dissociates as $[Pd(H_2O)_3Cl_3]Cl \cdot 3H_2O ightarrow [Pd(H_2O)_3Cl_3]^+ + Cl^-$. This gives $2$ ions $(i=2)$. Therefore,the correct formula is $[Pd(H_2O)_3Cl_3]Cl \cdot 3H_2O$.

Compound $PdCl_4 \cdot 6H_2O$ is a hydrated complex; $1 \ m$ aqueous solution of it has a freezing point of $269.28 \ K$. Assuming $100\%$ ionization of the complex,calculate the molecular formula of the complex ($K_f$ for water $= 1.86 \ K \ kg \ mol^{-1}$,freezing point of pure water $= 273 \ K$).

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