Question

The procedure in an experiment calls for a \(1.00 \mathrm{M}\) solution of KCl. A student prepares the solution by adding \(37.3 \mathrm{~g}\) to \(\mathrm{KCl}\) to \(500 \mathrm{~mL}\) of water. Does the resulting solution have the correct molarity? Explain.

Short answer

Yes, the solution is approximately 1.00 M, which is correct for the experiment.

Step-by-step solution

Calculate the Moles of KCl Needed

To prepare a 1.00 Molar (M) solution, we need to calculate the number of moles of KCl that are required for 500 mL (0.500 L) of solution. The formula for molarity is \( M = \frac{n}{V} \), where \( n \) is the number of moles and \( V \) is the volume in liters. Thus, \( n = M \times V \). For our solution: \( n = 1.00 \, M \times 0.500 \, L = 0.500 \, moles \).

Determine the Molar Mass of KCl

The molar mass of KCl is calculated as the sum of the atomic masses of potassium (K) and chlorine (Cl). The atomic masses are approximately 39.10 g/mol for K and 35.45 g/mol for Cl. Therefore, the molar mass: 39.10 g/mol + 35.45 g/mol = 74.55 g/mol.

Calculate the Mass of KCl Required

Using the molar mass, we determine the mass of KCl required for 0.500 moles. \( \text{Mass} = \text{molar mass} \times \text{number of moles} = 74.55 \, g/mol \times 0.500 \, moles = 37.275 \, g \).

Compare the Mass of KCl Used to the Required Mass

The student used 37.3 g of KCl to make the solution. This is very close to the calculated 37.275 g needed. The small difference is negligible in practical terms, meaning the molarity will be approximately 1.00 M.

Key concepts

Solution Preparation

Preparing a solution is a fundamental aspect of chemistry. When the instructions are to prepare a 1.00 M solution of KCl, it involves dissolving a precise amount of KCl in a specified volume of water. This process requires balancing the quantity of solute, KCl here, and the solvent, which is water, to achieve the desired molarity. The volume of the solution is crucial because molarity depends on it. For accuracy, use measuring tools like a balance for weighing and a volumetric flask for solution volumes. Ensuring the solute dissolves completely without leaving any residues is equally important for consistency.

Moles Calculation

Moles calculation is key in finding out how much solute is needed for a given molarity and volume. Molarity is the number of moles of solute per liter of solution and is represented by the formula \( M = \frac{n}{V} \).
To find the number of moles \( n \), rearrange the formula: \( n = M \times V \). If you need a 1.00 M solution and have 0.500 L of volume, you calculate the moles as \( 1.00 \times 0.500 = 0.500 \) moles.
This step is crucial because it ensures the solution will have the correct concentration when prepared properly.

Molar Mass

Understanding molar mass helps in connecting moles to grams, which are a more practical measurement for preparing solutions. The molar mass is the mass of one mole of a substance and is calculated by summing the atomic masses of its elements.
For KCl: Potassium (K) has an atomic mass of approximately 39.10 g/mol, while Chlorine (Cl) has 35.45 g/mol. Therefore, the molar mass of KCl is \( 39.10 + 35.45 = 74.55 \) g/mol.
This value is essential as it converts the needed moles to grams, allowing us to weigh and prepare the solution accurately.

Concentration Analysis

Concentration analysis ensures that the solution has been prepared correctly. It involves verifying that the measured and calculated values align to give the desired solution molarity.
In this scenario, the calculated mass of KCl required for 0.500 moles is \( 74.55 \times 0.500 = 37.275 \) g.
The student used 37.3 g, which nearly matches the calculated needed mass, confirming the target of 1.00 M molarity is achieved within a small variance. This slight difference is typically negligible and acceptable in most practical settings.

• Thorough stirring helps ensure even distribution of solute.
• Always cross-check your calculated values against the actual prepared amounts.

By following these steps, the concentration analysis confirms the success of the solution preparation.