A solution containing `2.675g` of `COCl_(3).6NH_(3)` (molar mass `=267.5gmol^(-1)`) is passed through a cation exchanger, The chloride ions obtined in solution were treated with excess of `AgNO_(3)` to give`4.78g` of `AgCl ("molar mass" =143.5gmol^(-1)`.The formula of the complex is (Atomic mass of `Ag=108 u`)

Options

A. `[Co(NH_(3))_(6)]Cl_(3)`
B. `[CoCl_(2)(NH_(3))_(4)]Cl`
C. `[CoCl_(3)(NH_(3))_(3)]`
D. `[CoCl(NH_(3))_(5)]Cl_(2)`

Explanation

To solve the problem step by step, we need to determine the formula of the complex COCl₃·6NH₃ based on the information provided.

### Step 1: Calculate the moles of COCl₃·6NH₃

Given:
- Mass of COCl₃·6NH₃ = 2.675 g
- Molar mass of COCl₃·6NH₃ = 267.5 g/mol

To find the number of moles, we use the formula:
\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \]
Substituting the values:
\[ \text{Moles of COCl}_3·6\text{NH}_3 = \frac{2.675 \text{ g}}{267.5 \text{ g/mol}} \approx 0.01 \text{ mol} \]

### Step 2: Calculate the moles of AgCl produced

Given:
- Mass of AgCl = 4.78 g
- Molar mass of AgCl = 143.5 g/mol

Using the same formula for moles:
\[ \text{Moles of AgCl} = \frac{4.78 \text{ g}}{143.5 \text{ g/mol}} \approx 0.0333 \text{ mol} \]

### Step 3: Determine the moles of Cl⁻ ions

From the reaction, each mole of COCl₃·6NH₃ produces 3 moles of Cl⁻ ions. Therefore, if we have 0.01 moles of COCl₃·6NH₃, the moles of Cl⁻ ions produced will be:
\[ \text{Moles of Cl}^- = 3 \times \text{Moles of COCl}_3·6\text{NH}_3 = 3 \times 0.01 \text{ mol} = 0.03 \text{ mol} \]

### Step 4: Relate the moles of Cl⁻ ions to AgCl

Since each mole of AgCl corresponds to one mole of Cl⁻ ions, the moles of AgCl produced (0.0333 mol) indicates that there are more Cl⁻ ions than what was produced from COCl₃·6NH₃. This means that the complex COCl₃·6NH₃ must contain more Cl⁻ ions than just the ones from the complex itself.

### Step 5: Determine the total number of Cl⁻ ions in the complex

Since we have 0.0333 moles of AgCl, and each mole of AgCl corresponds to one mole of Cl⁻, we can say:
\[ \text{Total moles of Cl}^- = 0.0333 \text{ mol} \]

### Step 6: Calculate the number of Cl⁻ ions in the complex

From the previous calculation, we found that 0.03 moles of Cl⁻ ions were produced from the complex. The additional Cl⁻ ions must come from the complex itself. Therefore, we can conclude that the complex has:
\[ \text{Total Cl}^- = 0.0333 \text{ mol} - 0.03 \text{ mol} = 0.0033 \text{ mol} \]

### Step 7: Determine the formula of the complex

Since the complex COCl₃·6NH₃ has 3 Cl⁻ ions, and we have established that the total number of Cl⁻ ions is 0.0333 mol, we can conclude that the complex must contain an additional Cl⁻ ion. Therefore, the formula of the complex is:
\[ \text{COCl}_4·6\text{NH}_3 \]

### Final Answer:
The formula of the complex is COCl₄·6NH₃.

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