Complex A has a composition of H_{12}O_{6}Cl_ | vedclass

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(C) The molar mass of the complex $CrCl_{3} \cdot 6H_{2}O$ is calculated as: $52 + (3 	imes 35.5) + (6 	imes 18) = 52 + 106.5 + 108 = 266.5 \ g/mol$

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$[Cr(H_{2}O)_{4}Cl_{2}]Cl \cdot 2H_{2}O$

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(C) The molar mass of the complex $CrCl_{3} \cdot 6H_{2}O$ is calculated as: $52 + (3 	imes 35.5) + (6 	imes 18) = 52 + 106.5 + 108 = 266.5 \ g/mol$.Conc. $H_{2}SO_{4}$ acts as a dehydrating agent and removes water molecules present outside the coordination sphere (lattice water).Let $x$ be the number of water molecules lost.Percentage mass loss $= \frac{x 	imes 18}{266.5} 	imes 100 = 13.5$.$x = \frac{13.5 	imes 266.5}{1800} \approx 2$.Since $2$ water molecules are lost,the complex must contain $2$ molecules of lattice water.The formula is $[Cr(H_{2}O)_{4}Cl_{2}]Cl \cdot 2H_{2}O$.

Complex $A$ has a composition of $H_{12}O_{6}Cl_{3}Cr$. If the complex on treatment with conc. $H_{2}SO_{4}$ loses $13.5\%$ of its original mass,the correct molecular formula of $A$ is :
[Given : atomic mass of $Cr = 52 \ amu$ and $Cl = 35.5 \ amu$]

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Complex $A$ has a composition of $H_{12}O_{6}Cl_{3}Cr$. If the complex on treatment with conc. $H_{2}SO_{4}$ loses $13.5\%$ of its original mass,the correct molecular formula of $A$ is :[Given : atomic mass of $Cr = 52 \ amu$ and $Cl = 35.5 \ amu$]