If the freezing point of a 0.01 m aqueous so | vedclass

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(A) The depression in freezing point is given by $\Delta T_{f} = i K_{f} m$. Given $\Delta T_{f} = 0 - (-0.0558) = 0.0558 \ K$,$K_{f} = 1.86 \ K \ kg

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(A) The depression in freezing point is given by $\Delta T_{f} = i K_{f} m$. Given $\Delta T_{f} = 0 - (-0.0558) = 0.0558 \ K$,$K_{f} = 1.86 \ K \ kg \ mol^{-1}$,and $m = 0.01 \ m$. Substituting the values: $0.0558 = i 	imes 1.86 	imes 0.01$. $i = \frac{0.0558}{0.0186} = 3$. Since the complex is a cobalt $(III)$ chloride-ammonia complex,it is of the form $[Co(NH_3)_xCl_y]Cl_z$. For $i = 3$,the complex dissociates into $3$ ions: $[Co(NH_3)_xCl_y]^{2+} + 2Cl^-$. This implies there are $2$ chloride ions outside the coordination sphere. Since the coordination number of $Co(III)$ is $6$,the formula is $[Co(NH_3)_5Cl]Cl_2$. Thus,the number of chloride ions inside the coordination sphere is $1$.

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