Complex Structure Determination

The spin-only magnetic moment is given as $3.87 \text{ BM}$. Using the formula $\mu_s = \sqrt{n(n+2)}$, where $n$ is the number of unpaired electrons:

$ \sqrt{n(n+2)} = 3.87 $

$ n(n+2) \approx 14.97 $

Solving for $n$, we find $n=3$. This corresponds to $\text{Cr}^{3+}$ ion (electronic configuration $3d^3$).

The conductivity corresponds to a $1:2$ electrolyte, meaning the complex dissociates into one cation and two anions in solution. Given the formula $\text{CrCl}_3\cdot 6\text{H}_2\text{O}$ and the $\text{Cr}^{3+}$ oxidation state, the complex structure must be $[\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2$. This structure dissociates as:

$ [\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]^{2+}(\text{aq}) + 2\text{Cl}^-(\text{aq}) $

Cation Exchange Analysis

A cation exchanger replaces cations in solution with $\text{H}^+$ ions. In this process, the complex cation $[Cr(\text{H}_2\text{O})_5\text{Cl}]^{2+}$ is retained by the resin.

The chloride ions ($\text{Cl}^-$) are anions and pass through the exchanger along with $\text{H}^+$ ions released from the resin.

Therefore, the solution obtained after passing through the cation exchanger contains the $2$ moles of $\text{Cl}^-$ ions that were initially associated with the complex cation.

AgCl Precipitation Calculation

The solution after cation exchange is treated with excess $\text{AgNO}_3$. The free $\text{Cl}^-$ ions precipitate as $\text{AgCl}$:

$ \text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s}) $

The molar mass of the complex $[\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2$ is calculated as:

Molar Mass $= \text{M}(\text{Cr}) + 5 \times (\text{M}(\text{H}_2\text{O})) + \text{M}(\text{Cl}) + 2 \times \text{M}(\text{Cl})$

Molar Mass $= 52 + 5 \times (2 \times 1 + 16) + 35.5 + 2 \times 35.5$

Molar Mass $= 52 + 5 \times 18 + 35.5 + 71 = 52 + 90 + 35.5 + 71 = 248.5 \text{ g mol}^{-1}$

The number of moles of the complex is:

$ \text{Moles of complex} = \frac{2.75 \text{ g}}{248.5 \text{ g mol}^{-1}} \approx 0.011066 \text{ mol} $

Since there are $2$ moles of $\text{Cl}^-$ ions per mole of complex available for precipitation:

$ \text{Moles of Cl}^- = 2 \times 0.011066 \text{ mol} \approx 0.022132 \text{ mol} $

The moles of $\text{AgCl}$ formed are equal to the moles of $\text{Cl}^-$:

$ \text{Moles of AgCl} = 0.022132 \text{ mol} $

The molar mass of $\text{AgCl}$ is $\text{M}(\text{Ag}) + \text{M}(\text{Cl}) = 108 + 35.5 = 143.5 \text{ g mol}^{-1}$.

The mass of $\text{AgCl}$ formed is:

$ \text{Mass of AgCl} = 0.022132 \text{ mol} \times 143.5 \text{ g mol}^{-1} \approx 3.1767 \text{ g} $

Final Answer Determination

Rounding the calculated mass of $\text{AgCl}$ to the nearest integer gives $3 \text{ g}$.