Problem 33 - Chemistry An Atoms First Approach

Question

Which of the following solutions of strong electrolytes contains the largest number of moles of chloride ions: \(100.0 \mathrm{mL}\) of \(0.30 \mathrm{M} \mathrm{AlCl}_{3}, 50.0 \mathrm{mL}\) of \(0.60 \mathrm{M} \mathrm{MgCl}_{2},\) or \(200.0 \mathrm{mL}\) of \(0.40 M\) NaCl?

Step-by-Step Solution

Answer

The 100.0 mL of 0.30 M AlCl3 solution contains the largest number of moles of chloride ions, with a total of 0.090 mol of Cl- ions.

Step 1: List given information and formula to be used.

We have been given the following information: 1. 100.0 mL of a 0.30 M AlCl3 solution 2. 50.0 mL of a 0.60 M MgCl2 solution 3. 200.0 mL of a 0.40 M NaCl solution We need to find the number of moles of Cl- in each solution. The formula to calculate the moles is: moles = volume × molarity × ions in the solution

Step 2: Calculate moles of Cl- in AlCl3 solution.

First, we will find the number of moles of Cl- in the 100.0 mL of 0.30 M AlCl3 solution. moles of Cl- in AlCl3 = 100.0 mL × 0.30 M × 3 since there are 3 Cl- ions in AlCl3. Convert mL to L: 100.0 mL = 0.100 L moles of Cl- in AlCl3 = 0.100 L × 0.30 M × 3 moles of Cl- in AlCl3 = 0.090 mol

Step 3: Calculate moles of Cl- in MgCl2 solution.

Now, we will find the number of moles of Cl- in the 50.0 mL of 0.60 M MgCl2 solution. moles of Cl- in MgCl2 = 50.0 mL × 0.60 M × 2 since there are 2 Cl- ions in MgCl2. Convert mL to L: 50.0 mL = 0.050 L moles of Cl- in MgCl2 = 0.050 L × 0.60 M × 2 moles of Cl- in MgCl2 = 0.060 mol

Step 4: Calculate moles of Cl- in NaCl solution.

Lastly, we will find the number of moles of Cl- in the 200.0 mL of 0.40 M NaCl solution. moles of Cl- in NaCl = 200.0 mL × 0.40 M × 1 since there is 1 Cl- ion in NaCl. Convert mL to L: 200.0 mL = 0.200 L moles of Cl- in NaCl = 0.200 L × 0.40 M × 1 moles of Cl- in NaCl = 0.080 mol

Step 5: Compare moles and find the largest quantity of chloride ions.

Now, let's compare the moles of chloride ions found in each solution: 1. AlCl3: 0.090 mol 2. MgCl2: 0.060 mol 3. NaCl: 0.080 mol The largest number of moles of chloride ions is in the 100.0 mL of 0.30 M AlCl3 solution, which has 0.090 mol of Cl- ions.

Key Concepts

Understanding Molarity

Molarity is one of the core concepts in chemistry, especially when analyzing solution concentrations. It represents the number of moles of a solute that is present in one liter of solution. The unit for molarity is Molar, often abbreviated as M. Essentially, molarity can be thought of as a way to express the strength or concentration of a solution.

Molarity is calculated using the formula:
\( \text{Molarity} (M) = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \)

When comparing solutions to determine which one has a higher concentration of a certain ion or component, understanding molarity is crucial. For example, in our exercise, to find out which solution has the largest number of moles of chloride ions, we need to know the molarity of each solution and realize that it directly affects the amount of ions.

Calculating Moles of Ions

The process of calculating moles of ions in a solution requires a clear understanding of stoichiometry and the chemical formula of the solute. For instance, when a compound like Aluminum Chloride (\(\mathrm{AlCl}_{3}\)) dissolves in water, it dissociates into one Aluminum ion (\(\mathrm{Al}^{3+}\)) and three Chloride ions (\(\mathrm{Cl}^{-}\)). Thus, the number of moles of Chloride ions will be three times the number of moles of \(\mathrm{AlCl}_{3}\) used.

To calculate the moles of ions, you can use the formula:
\( \text{Moles of ions} = \text{Volume (L)} \times \text{Molarity (M)} \times \text{Number of ions per formula unit} \)

It's essential to ensure the correct stoichiometric ratio is used. In our exercise's case, we multiply by 3 for \(\mathrm{AlCl}_{3}\), by 2 for \(\mathrm{MgCl}_{2}\), and by 1 for NaCl since these are the number of Chloride ions present per formula unit of each compound.

Volume Conversion in Solutions

Volume conversion is a vital skill when working with solutions, especially when volumes are provided in different units like milliliters (mL) and liters (L). To perform accurate calculations, volumes typically need to be converted to liters since molarity is defined per liter of solution.

The conversion between milliliters and liters is straightforward:
\( 1 \text{ L} = 1,000 \text{ mL} \)
\( \text{Volume (L)} = \frac{\text{Volume (mL)}}{1,000} \)

It's essential not to overlook this step, as failing to convert volume could lead to incorrect calculations of molarity and moles of ions, as demonstrated in our exercise where all the volumes were initially given in milliliters and needed to be converted to liters to find out the number of moles of chloride ions accurately.