`A` chloride mixture is prepared by grinding together pure `BaCl_(2).2H_(2)O, KCl` and `NaCl`. What is the smallest and largest volume of `0.15 M AgNO_(3)` solution that may be used for complete precipitation of chloride from a `0.3 g` sample of the mixture which may contain any one or all of the constituents ?

Explanation

To solve the problem, we need to determine the smallest and largest volumes of 0.15 M AgNO₃ solution required to completely precipitate chloride ions from a 0.3 g sample of a mixture containing BaCl₂·2H₂O, KCl, and NaCl.

### Step-by-Step Solution:

1. **Identify the Molar Masses**:
- Molar mass of BaCl₂·2H₂O:
- Ba = 137.33 g/mol
- Cl = 35.45 g/mol × 2 = 70.90 g/mol
- H₂O = 18.02 g/mol × 2 = 36.04 g/mol
- Total = 137.33 + 70.90 + 36.04 = 244.27 g/mol
- Molar mass of KCl:
- K = 39.10 g/mol
- Cl = 35.45 g/mol
- Total = 39.10 + 35.45 = 74.55 g/mol
- Molar mass of NaCl:
- Na = 22.99 g/mol
- Cl = 35.45 g/mol
- Total = 22.99 + 35.45 = 58.44 g/mol

2. **Calculate the Moles of Each Constituent**:
- For the smallest volume (using the highest molar mass constituent, BaCl₂·2H₂O):
\[ \text{Moles of BaCl₂·2H₂O} = \frac{0.3 \text{ g}}{244.27 \text{ g/mol}} \approx 0.00123 \text{ moles} \]
- For the largest volume (using the lowest molar mass constituent, NaCl):
\[ \text{Moles of NaCl} = \frac{0.3 \text{ g}}{58.44 \text{ g/mol}} \approx 0.00513 \text{ moles} \]

3. **Determine the Moles of AgNO₃ Required**:
- Each mole of chloride requires one mole of AgNO₃ for precipitation.
- For BaCl₂·2H₂O (which provides 2 moles of Cl):
\[ \text{Moles of AgNO₃} = 2 \times 0.00123 \text{ moles} \approx 0.00246 \text{ moles} \]
- For NaCl (which provides 1 mole of Cl):
\[ \text{Moles of AgNO₃} = 0.00513 \text{ moles} \]

4. **Calculate the Volume of AgNO₃ Solution Required**:
- Using the formula \( \text{Volume} = \frac{\text{Moles}}{\text{Concentration}} \):
- For the smallest volume (from BaCl₂·2H₂O):
\[ \text{Volume of AgNO₃} = \frac{0.00246 \text{ moles}}{0.15 \text{ M}} \approx 0.0164 \text{ L} = 16.4 \text{ mL} \]
- For the largest volume (from NaCl):
\[ \text{Volume of AgNO₃} = \frac{0.00513 \text{ moles}}{0.15 \text{ M}} \approx 0.0342 \text{ L} = 34.2 \text{ mL} \]

### Final Answer:
- **Smallest Volume of AgNO₃**: 16.4 mL
- **Largest Volume of AgNO₃**: 34.2 mL