Question

5.33 g of \(CrCl_3 \cdot 6H_2O\), which is a 1:3 electrolyte, is dissolved in water and is passed through a cation exchanger. The chloride ions in the eluted solution, on treatment with \(AgNO_3\), results in 8.61 g of \(AgCl\). The ratio of moles of complex reacted and moles of \(AgCl\) formed is ______ \(\times 10^{-2}\). (Nearest integer)} \[ [Molar\ mass: Cr = 52,\ Ag = 108,\ Cl = 35.5,\ H = 1,\ O = 16] \]

Answer 1

Correct Answer: 25

Step 1: Understanding the Concept:
The term "1 : 3 electrolyte" for the complex \(CrCl_3 \cdot 6H_2O\) indicates that its coordination formula is \([Cr(H_2O)_6]Cl_3\).
When this complex is passed through a cation exchanger, the complex cation \([Cr(H_2O)_6]^{3+}\) is trapped by the resin and replaced by \(H^+\) ions.
The chloride ions (\(Cl^-\)) are not exchanged and remain in the eluted solution.
These ions then react with silver nitrate (\(AgNO_3\)) to form a precipitate of silver chloride (\(AgCl\)).
Step 2: Key Formula or Approach:
1. Molar Mass of \([Cr(H_2O)_6]Cl_3 = 52 + (6 \times 18) + (3 \times 35.5) = 52 + 108 + 106.5 = 266.5 \text{ g/mol}\).
2. Molar Mass of \(AgCl = 108 + 35.5 = 143.5 \text{ g/mol}\).
3. Number of moles \(n = \frac{\text{given mass}}{\text{molar mass}}\).
Step 3: Detailed Explanation:
First, calculate the number of moles of the complex used:
\[ n_{\text{complex}} = \frac{5.33}{266.5} = 0.02 \text{ mol} \]
Next, calculate the number of moles of \(AgCl\) precipitated:
\[ n_{AgCl} = \frac{8.61}{143.5} = 0.06 \text{ mol} \]
Now, find the ratio of moles of complex reacted to moles of \(AgCl\) formed:
\[ \text{Ratio} = \frac{n_{\text{complex}}}{n_{AgCl}} = \frac{0.02}{0.06} = \frac{1}{3} \approx 0.3333 \]
The question asks for the answer in the form of \(\text{Value} \times 10^{-2}\):
\[ 0.3333 = 33.33 \times 10^{-2} \]
Rounding to the nearest integer, the value is 33.
Step 4: Final Answer:
The ratio is \(33 \times 10^{-2}\).