Chloride samples are prepared for analysis by using NaCl. KCl and `NH_(4)Cl` separately or as mixture. What minimum volume (in mL) of 5% by mass `AgNO_(3)` solution (sp. Gravity 1.19 g/mL) must be added to a sample of 0.1498 g in order to ensure precipitation of chloride in every possible case?

Explanation

To solve the problem, we need to determine the minimum volume of a 5% by mass AgNO₃ solution required to ensure complete precipitation of chloride ions from a sample weighing 0.1498 g.

### Step-by-Step Solution:

1. **Determine the mass of chloride in the sample:**
- The sample can contain NaCl, KCl, and NH₄Cl. We will consider the maximum possible chloride content, which is from NH₄Cl.
- The molar mass of NH₄Cl is approximately 53.5 g/mol.

2. **Calculate the moles of NH₄Cl in the sample:**
\[ \text{Moles of NH₄Cl} = \frac{\text{mass of sample}}{\text{molar mass of NH₄Cl}} = \frac{0.1498 \, \text{g}}{53.5 \, \text{g/mol}} \approx 0.00280 \, \text{mol} \]

3. **Determine the moles of AgNO₃ required for precipitation:**
- The reaction is:
\[ \text{AgNO}_3 + \text{NH}_4\text{Cl} \rightarrow \text{AgCl} + \text{NH}_4\text{NO}_3 \]
- From the reaction, 1 mole of NH₄Cl reacts with 1 mole of AgNO₃. Therefore, the moles of AgNO₃ required are equal to the moles of NH₄Cl:
\[ \text{Moles of AgNO₃ required} = 0.00280 \, \text{mol} \]

4. **Calculate the mass of AgNO₃ required:**
- The molar mass of AgNO₃ is approximately 169.87 g/mol.
\[ \text{Mass of AgNO₃} = \text{moles} \times \text{molar mass} = 0.00280 \, \text{mol} \times 169.87 \, \text{g/mol} \approx 0.476 \, \text{g} \]

5. **Determine the volume of 5% AgNO₃ solution needed:**
- A 5% (w/v) solution means 5 g of AgNO₃ in 100 mL of solution.
- The density of the solution is given as 1.19 g/mL, which means 100 mL of solution weighs 119 g.
- The mass of AgNO₃ in 1 mL of solution is:
\[ \text{Mass of AgNO₃ in 1 mL} = \frac{5 \, \text{g}}{100 \, \text{mL}} = 0.05 \, \text{g/mL} \]
- To find the volume (V) of the solution needed to get 0.476 g of AgNO₃:
\[ V = \frac{\text{mass of AgNO₃ required}}{\text{mass of AgNO₃ in 1 mL}} = \frac{0.476 \, \text{g}}{0.05 \, \text{g/mL}} \approx 9.52 \, \text{mL} \]

6. **Final Calculation:**
- Since we need to ensure complete precipitation, we round up to the nearest whole number. Thus, we will need at least 10 mL of the 5% AgNO₃ solution.

### Final Answer:
The minimum volume of 5% by mass AgNO₃ solution required is **10 mL**.