A compound Cr(H₂O)₆⋅Cl₃ show conductance similar to 1 : 2 electrolyte in aq. solution. 9.6 g of this complex is passed through a cation exchanger then excess of AgNO3 solution is added. Find mass of AgCl precipitated in gram?

[Molar mas of Cr = 52, Cl = 35.5]

The correct answer is 10.

Step-by-step Explanation:

1. Determine the structure of the coordination complex:
The compound has the empirical formula Cr(H₂O)₆·Cl₃.
It is given that the compound behaves as a 1 : 2 electrolyte in aqueous solution. A 1 : 2 electrolyte dissociates to yield 1 cation and 2 anions (a total of 3 ions).
This indicates that two chloride ions reside outside the coordination sphere as counter-ions, while one chloride ion is coordinated directly to the chromium center.
Therefore, the structural formula of the complex is:
$\left[\mathrm{Cr}\right(H_{2}O{)}_5\mathrm{Cl}]{\mathrm{Cl}}_2\cdot H_{2}O$
When dissolved in water, it dissociates as follows:
$\left[\mathrm{Cr}\right(H_{2}O{)}_5\mathrm{Cl}]{\mathrm{Cl}}_2\cdot H_{2}O(\mathrm{aq})\to {\left[\mathrm{Cr}\right(H_{2}O{)}_5\mathrm{Cl}]}^{2+}(\mathrm{aq})+2{\mathrm{Cl}}^{-}(\mathrm{aq})$
Thus, each mole of the complex yields 2 moles of free chloride ions (${\mathrm{Cl}}^{-}$) in solution.

2. Calculate the molar mass of the complex:
Using the given molar masses:
Molar mass of Cr = 52 g/mol
Molar mass of Cl = 35.5 g/mol
Molar mass of H = 1 g/mol, O = 16 g/mol (so H₂O = 18 g/mol)
$\text{Molar Mass of}\mathrm{Cr}(H_{2}O{)}_6{\mathrm{Cl}}_3=52+(6\times 18)+(3\times 35.5)$
$=52+108+106.5=266.5\text{g/mol}$

3. Calculate the number of moles of the complex:
We are given 9.6 g of the complex:
$\text{Moles of complex}=\frac{9.6}{266.5}\approx 0.036\text{mol}$

4. Analyze the effect of the cation exchanger:
When the solution is passed through a cation exchanger, the complex cations ${\left[\mathrm{Cr}\right(H_{2}O{)}_5\mathrm{Cl}]}^{2+}$ are exchanged with $H^{+}$ ions.
The chloride anions (${\mathrm{Cl}}^{-}$) do not undergo exchange and flow out into the effluent.
Therefore, the number of moles of ${\mathrm{Cl}}^{-}$ in the effluent remains unchanged:
$\text{Moles of}{\mathrm{Cl}}^{-}=2\times \text{Moles of complex}$
$=2\times 0.036\text{mol}=0.072\text{mol}$

5. Calculate the mass of AgCl precipitated:
When excess AgNO₃ is added to the effluent, all free chloride ions react to precipitate AgCl:
${\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\to \mathrm{AgCl}\left(s\right)$
Hence, the moles of AgCl precipitated are equal to the moles of chloride ions:
$\text{Moles of AgCl precipitated}=0.072\text{mol}$
The molar mass of AgCl is:
$\text{Molar mass of AgCl}=108\left(\mathrm{Ag}\right)+35.5\left(\mathrm{Cl}\right)=143.5\text{g/mol}$
Thus, the mass of AgCl precipitated is:
$\text{Mass of AgCl}=0.072\text{mol}\times 143.5\text{g/mol}\approx 10.33\text{g}$
Rounding to the nearest integer, the mass of AgCl precipitated is 10 grams.