A 0.001 molal solution of [Pt(NH_3)_4Cl_4] i | vedclass

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(B) The formula for freezing point depression is $\Delta T_f = i 	imes K_f 	imes m$. Given $\Delta T_f = 0.0054 \ ^oC$,$K_f = 1.80 \ K \ kg \ mol^{-

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$[Pt(NH_3)_4Cl_2]Cl_2$

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(B) The formula for freezing point depression is $\Delta T_f = i 	imes K_f 	imes m$. Given $\Delta T_f = 0.0054 \ ^oC$,$K_f = 1.80 \ K \ kg \ mol^{-1}$,and $m = 0.001 \ molal$. 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 must dissociate into $3$ ions in solution. The complex $[Pt(NH_3)_4Cl_2]Cl_2$ dissociates as $[Pt(NH_3)_4Cl_2]^{2+} + 2Cl^-$,giving a total of $3$ ions. Therefore,the correct formulation is $[Pt(NH_3)_4Cl_2]Cl_2$.

$A$ $0.001 \ molal$ solution of $[Pt(NH_3)_4Cl_4]$ in water has a freezing point depression of $0.0054 \ ^oC$. If $K_f$ for water is $1.80 \ K \ kg \ mol^{-1}$,the correct formulation for the above complex is:

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