CrCl_3 cdot xNH_3 can exist as a complex. A 0 | vedclass

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(C) Given: $\Delta T_f = 0.558^{\circ} C$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,$m = 0.1 \ m$. Using the formula for depression in freezing point: $\Delta

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$[Cr(NH_3)_5Cl]Cl_2$

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(C) Given: $\Delta T_f = 0.558^{\circ} C$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,$m = 0.1 \ m$. Using the formula for depression in freezing point: $\Delta T_f = i 	imes K_f 	imes m$. Substituting the values: $0.558 = i 	imes 1.86 	imes 0.1$. Solving for $i$: $i = 0.558 / 0.186 = 3$. Since the van't Hoff factor $i = 3$,the complex dissociates into $3$ ions. For the complex $[Cr(NH_3)_5Cl]Cl_2$,the dissociation is $[Cr(NH_3)_5Cl]Cl_2 ightarrow [Cr(NH_3)_5Cl]^{2+} + 2Cl^{-}$,which yields $1 + 2 = 3$ ions. Thus,the complex is $[Cr(NH_3)_5Cl]Cl_2$.

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