The elevation of boiling point of 0.10 m aqu | vedclass

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(B) The elevation in boiling point is given by $\Delta T_{b} = i 	imes K_{b} 	imes m$.For $0.05 \ m$ $CaCl_{2}$ solution,$i = 3$ (since $CaCl_{2}

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$5$

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(B) The elevation in boiling point is given by $\Delta T_{b} = i 	imes K_{b} 	imes m$.For $0.05 \ m$ $CaCl_{2}$ solution,$i = 3$ (since $CaCl_{2} ightarrow Ca^{2+} + 2Cl^{-}$). Thus,$\Delta T_{b(CaCl_{2})} = 3 	imes 0.05 	imes K_{b} = 0.15 	imes K_{b}$.For $0.10 \ m$ $CrCl_{3} \cdot xNH_{3}$ solution,$\Delta T_{b(complex)} = i 	imes 0.10 	imes K_{b}$.Given $\Delta T_{b(complex)} = 2 	imes \Delta T_{b(CaCl_{2})}$,we have $i 	imes 0.10 	imes K_{b} = 2 	imes (0.15 	imes K_{b}) = 0.30 	imes K_{b}$.Therefore,$i = 3$.Since the complex $[Cr(NH_{3})_{x}Cl_{3}]$ ionizes to give $i = 3$ ions,it must dissociate as $[Cr(NH_{3})_{x}Cl_{3-y}]^{y+} + yCl^{-}$,where $1 + y = 3$,so $y = 2$.This means the complex is $[Cr(NH_{3})_{x}Cl]Cl_{2}$.Given the coordination number of $Cr$ is $6$,the number of ligands is $x + 1 = 6$,which gives $x = 5$.

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