5.33 mg of salt [`Cr(H_(2)O)_(5)Cl].Cl_(2).H_(2)O` is treated with excess of `AgNO_(3)`(aq) then mass of AgCl precipitate obtained will be :
Given : [Cr = 52, Cl = 35.5 ,Ag = 108 ]

Answer

To solve the problem, we need to find the mass of AgCl precipitate obtained when 5.33 mg of the salt \[Cr(H_2O)_5Cl \cdot Cl_2 \cdot H_2O\] is treated with excess AgNO3. Here are the steps to arrive at the solution:

### Step 1: Calculate the molar mass of the salt \[Cr(H_2O)_5Cl \cdot Cl_2 \cdot H_2O\]

1. **Identify the components of the salt:**
- Chromium (Cr): 1 atom
- Water (H2O): 5 molecules
- Chlorine (Cl): 3 atoms (1 from \[Cl\] and 2 from \[Cl_2\])
- Water of crystallization: 1 molecule

2. **Calculate the molar mass:**
- Molar mass of Cr = 52 g/mol
- Molar mass of H2O = 18 g/mol (since H = 1 and O = 16)
- Molar mass of Cl = 35.5 g/mol

\[ \text{Molar mass of the salt} = 52 + (5 \times 18) + (3 \times 35.5) \]
\[ = 52 + 90 + 106.5 = 248.5 \text{ g/mol} \]

### Step 2: Convert the mass of the salt from mg to g

\[ 5.33 \text{ mg} = 5.33 \times 10^{-3} \text{ g} \]

### Step 3: Calculate the number of moles of the salt

Using the formula for moles:

\[ \text{Moles of salt} = \frac{\text{mass}}{\text{molar mass}} = \frac{5.33 \times 10^{-3} \text{ g}}{248.5 \text{ g/mol}} \]

Calculating this gives:

\[ \text{Moles of salt} \approx 0.000214 \text{ mol} \]

### Step 4: Determine the moles of AgCl produced

From the reaction, we know that 1 mole of the salt produces 2 moles of AgCl. Therefore:

\[ \text{Moles of AgCl} = 2 \times \text{Moles of salt} = 2 \times 0.000214 \text{ mol} \approx 0.000428 \text{ mol} \]

### Step 5: Calculate the mass of AgCl precipitate

The molar mass of AgCl is:

\[ \text{Molar mass of AgCl} = 108 + 35.5 = 143.5 \text{ g/mol} \]

Now, calculate the mass of AgCl:

\[ \text{Mass of AgCl} = \text{Moles of AgCl} \times \text{Molar mass of AgCl} = 0.000428 \text{ mol} \times 143.5 \text{ g/mol} \]

Calculating this gives:

\[ \text{Mass of AgCl} \approx 0.0615 \text{ g} = 61.5 \text{ mg} \]

### Final Answer:

The mass of AgCl precipitate obtained will be approximately **61.5 mg**.

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