Question

A sample containing \(0.221 \mathrm{g} \mathrm{Cl}^{-}\) is dissolved in \(50.0 \mathrm{mL}\) water. a. How many moles of \(\mathrm{Cl}^{-}\) ion are in the solution? b. What is the molarity of the \(\mathrm{Cl}^{-}\) ion in the solution? \(\left(M_{\mathrm{Cl}^{-}}=n_{\mathrm{Cl}^{-}} / V_{\text {soln }}\right)\)

Short answer

a. There are 0.00623 moles of Cl⁻ in the sample. b. The molarity of the Cl⁻ ions in the solution is 0.1246 mol/L.

Step-by-step solution

Find the molecular weight of \(\mathrm{Cl}^{-}\)

To calculate the number of moles of Cl⁻, we first need to find the molecular weight of Cl⁻. We know that chlorine (Cl) has an atomic weight of 35.45 g/mol. Since Cl⁻ has just one chlorine atom, its molecular weight is also 35.45 g/mol.

Calculate the number of moles of \(\mathrm{Cl}^{-}\) in the sample

Now that we know the molecular weight of Cl⁻, we can calculate the number of moles in our sample using the given mass (0.221 g) and the molar mass (35.45 g/mol). Number of moles = (mass of Cl⁻) / (molar mass of Cl⁻) = (0.221 g) / (35.45 g/mol) Number of moles = 0.00623 mol (rounded to 5 decimal places) So, there are 0.00623 moles of Cl⁻ in the sample.

Convert the volume of the solution to liters

We are given that the volume of the solution is 50.0 mL. To calculate the molarity of the solution, we need the volume in liters. To convert from mL to L, we simply divide by 1000: Volume (L) = Volume (mL) / 1000 = 50.0 mL / 1000 = 0.05 L

Calculate the molarity of the \(\mathrm{Cl}^{-}\) in the solution

We now have the number of moles of Cl⁻ and the volume of the solution in liters. To find the molarity of Cl⁻, we use the formula: Molarity = (number of moles) / (volume of the solution in L) Molarity of Cl⁻ = (0.00623 mol) / (0.05 L) Molarity of Cl⁻ = 0.1246 mol/L (rounded to 4 decimal places) So, the molarity of the Cl⁻ ions in the solution is 0.1246 mol/L.

Key concepts

Molecular Weight Calculation

Molecular weight is essential in chemistry, especially for converting between mass and moles, which helps in various calculations. For an ion like chloride (\( \mathrm{Cl}^{-} \)), we take into account the atomic weight of the element. Chlorine has an atomic weight of 35.45 g/mol according to the periodic table. This value, called the molar mass, is crucial for our calculations.

To find the molecular weight of \( \mathrm{Cl}^{-} \), recognize that its molecular weight is the same as its atomic weight since it is only a single atom ion. This makes the molecular weight of \( \mathrm{Cl}^{-} \) 35.45 g/mol. It's important not to mix up molecular weight with formula weight, as the latter is for compounds, which have more than one atom.

• Atomic Weight for Cl: 35.45 g/mol
• Molecular Weight for \( \mathrm{Cl}^{-} \): 35.45 g/mol

Chloride Ion Concentration

Knowing the concentration of an ion in a solution is necessary for understanding a solution's reactivity and behavior. The concentration is often measured as molarity, defined as moles of solute per liter of solution.

To find the moles of \( \mathrm{Cl}^{-} \) in the solution, use the formula: \[\text{Number of moles} = \frac{\text{mass of } \mathrm{Cl}^{-}}{\text{molecular weight of } \mathrm{Cl}^{-}}\]Given the sample mass of \(0.221 \text{ g}\) and molar mass (35.45 \text{ g/mol}), divide to find the moles:

• Number of Moles = \(\frac{0.221 g}{35.45 g/mol} \approx 0.00623\) mol

To calculate the molarity, determine the total volume of the solution in liters, then apply:\[\text{Molarity} = \frac{\text{Number of moles}}{\text{Volume in liters}}\]

• Molarity = \(\frac{0.00623 \, ext{mol}}{0.05 \, ext{L}} = 0.1246 \, \text{mol/L}\)

This gives the concentration of \( \mathrm{Cl}^{-} \) in the solution as 0.1246 mol/L.

Solution Volume Conversion

Converting the volume of a solution from milliliters to liters is a crucial step in molarity calculations. Molarity uses liters as the standard unit because it provides a clearer measure of concentration.

Conversion is straightforward: divide the volume in milliliters by 1000 to obtain the volume in liters. For the given problem, the dissolution occurs in 50.0 mL of water:

• Volume in Liters = \( \frac{50.0 \, \text{mL}}{1000} = 0.05 \, \text{L}\).

Once converted, this volume can be used in further calculations to determine solution concentration. Remember:

• 1 Liter = 1000 mL
• Always ensure the unit conversion is correct for accurate calculation.

Converting correctly ensures an accurate understanding of concentration and makes your calculations valid across various chemistry applications.