Question #44561

A coordination compound with the empirical formula $\ce{CrCl3 \cdot 6H2O}$ exists as three distinct hydrate isomers (A, B, and C). At $298 \text{ K}$, the limiting molar conductivity ($\Lambda_m^\circ$) of isomer B in aqueous solution is found to be $258.0 \text{ S cm}^2 \text{ mol}^{-1}$.

Given the limiting ionic conductivities at $298 \text{ K}$: $\lambda^\circ(\ce{Cl^-}) = 76.3 \text{ S cm}^2 \text{ mol}^{-1}$ $\lambda^\circ([\ce{Cr(H2O)6}]^{3+}) = 156.0 \text{ S cm}^2 \text{ mol}^{-1}$ $\lambda^\circ([\ce{Cr(H2O)5Cl}]^{2+}) = 105.4 \text{ S cm}^2 \text{ mol}^{-1}$ $\lambda^\circ([\ce{Cr(H2O)4Cl2}]^{+}) = 53.7 \text{ S cm}^2 \text{ mol}^{-1}$

A $200 \text{ mL}$ portion of a $0.05 \text{ M}$ aqueous solution of isomer B is treated with excess $\ce{AgNO3(aq)}$. Separately, a $0.01 \text{ mol}$ solid sample of isomer B is placed in a desiccator over concentrated $\ce{H2SO4}$ until a constant weight is achieved.

What is the mass of the $\ce{AgCl}$ precipitate formed, and what is the mass loss due to dehydration of the solid sample, respectively? (Given molar masses in $\text{g mol}^{-1}$: $\ce{Ag} = 108.0$, $\ce{Cl} = 35.5$, $\ce{H2O} = 18.0$)