A 0.3 g sample is suspected to contain either pure NaCl or KCl or NH4Cl or a mixture of any of these in any proportion by mass or a mixture of all of these mixed in any proportion by mass.

A solution of AgNO3 is provided which has density of 1.05 g/CC containing 5% AgNO3 by mass. Calculate the minimum volume of this solution required to precipitate all the Cl- ions from the samples as AgCl.

To determine the minimum volume of the AgNO3 solution required to precipitate all the Cl^- ions from the given sample, we first need to identify the total amount of Cl^- ions present in the 0.3 g sample. This involves calculating the maximum possible mass of Cl^- ions based on the different compounds that could be present: NaCl, KCl, and NH4Cl.

Step 1: Calculate the Maximum Mass of Cl^- Ions

Each of the compounds contributes Cl^- ions in different proportions:

• NaCl: The molar mass of NaCl is approximately 58.44 g/mol. The mass fraction of Cl in NaCl is:

Mass fraction of Cl in NaCl = (35.45 g/mol) / (58.44 g/mol) ≈ 0.606

• KCl: The molar mass of KCl is about 74.55 g/mol. The mass fraction of Cl in KCl is:

Mass fraction of Cl in KCl = (35.45 g/mol) / (74.55 g/mol) ≈ 0.476

• NH4Cl: The molar mass of NH4Cl is around 53.49 g/mol. The mass fraction of Cl in NH4Cl is:

Mass fraction of Cl in NH4Cl = (35.45 g/mol) / (53.49 g/mol) ≈ 0.662

To find the maximum possible mass of Cl^- ions in the 0.3 g sample, we will consider the scenario where the sample is entirely composed of each compound:

• If the sample is NaCl:

  Mass of Cl^- = 0.3 g × 0.606 ≈ 0.1818 g

• If the sample is KCl:

  Mass of Cl^- = 0.3 g × 0.476 ≈ 0.1428 g

• If the sample is NH4Cl:

  Mass of Cl^- = 0.3 g × 0.662 ≈ 0.1986 g

The maximum mass of Cl^- ions occurs when the sample is NH4Cl, which is approximately 0.1986 g.

Step 2: Calculate the Moles of Cl^- Ions

Next, we convert the mass of Cl^- ions to moles:

Moles of Cl^- = Mass of Cl^- / Molar mass of Cl = 0.1986 g / 35.45 g/mol ≈ 0.0056 mol

Step 3: Determine the Amount of AgNO3 Required

The precipitation reaction between Cl^- ions and AgNO3 can be represented as:

Ag⁺ + Cl^- → AgCl (s)

This indicates a 1:1 molar ratio between Ag⁺ and Cl^-. Therefore, the moles of AgNO3 required will also be 0.0056 mol.

Step 4: Calculate the Mass of AgNO3 Required

The molar mass of AgNO3 is approximately 169.87 g/mol. Thus, the mass of AgNO3 needed is:

Mass of AgNO3 = Moles of AgNO3 × Molar mass of AgNO3 = 0.0056 mol × 169.87 g/mol ≈ 0.951 g

Step 5: Determine the Volume of AgNO3 Solution Required

The AgNO3 solution has a density of 1.05 g/cm³ and contains 5% AgNO3 by mass. Therefore, in 100 g of this solution, there are 5 g of AgNO3.

To find the volume of the solution needed to obtain 0.951 g of AgNO3, we can set up the following proportion:

Volume of solution = (Mass of AgNO3 required / Mass percentage of AgNO3) × 100

Mass of solution needed = (0.951 g / 0.05) = 19.02 g

Now, using the density to find the volume:

Volume = Mass / Density = 19.02 g / 1.05 g/cm³ ≈ 18.13 cm³

Final Calculation

Thus, the minimum volume of the AgNO3 solution required to precipitate all the Cl^- ions from the sample is approximately 18.13 cm³.