Problem 60 - Chemistry: The Central Science

Question

(a) Suppose you prepare \(500 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) solution of some salt and then spill some of it. What happens to the concentration of the solution left in the container? (b) Suppose you prepare \(500 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) aqueous solution of some salt and let it sit out, uncovered, for a long time, and some water evaporates. What happens to the concentration of the solution left in the container? (c) \(A\) certain volume of a \(0.50 \mathrm{M}\) solution contains \(4.5 \mathrm{~g}\) of a salt. What mass of the salt is present in the same volume of a \(2.50 \mathrm{M}\) solution?

Step-by-Step Solution

Answer

(a) The concentration of the solution left in the container remains unchanged. (b) The concentration of the solution will increase after the water has evaporated. (c) The mass of salt in the same volume of a 2.50 M solution is 22.5 g.

Step 1: Understand the relationship between concentration and volume

The concentration of a solution is defined as the amount of solute (in this case, salt) divided by the amount of solution (the total volume). So when the volume of the solution changes, the concentration will be affected. In this problem, we are spilling some of the 0.10 M solution; thus, the volume is decreasing.

Step 2: Analyze the change in concentration

When we spill some of the solution, the amount of salt present in the container remains the same, only the total volume of the solution has been reduced. Since the mass of the solute (salt) stays the same, and the volume of the solution is decreased, the concentration of solution left in the container will remain unchanged. #(b) Effect of water evaporation on concentration#

Step 1: Understand the relationship between concentration and water evaporation

In this problem, water from the aqueous solution evaporates, meaning that the volume of the solvent (water) will decrease, but the mass of the solute (salt) will remain the same.

Step 2: Analyze the change in concentration

Since the volume of the solvent is decreasing and the mass of the solute remains constant, the concentration of the solution will increase after the water has evaporated. #(c) Finding the mass of salt in the new solution#

Step 1: Calculate moles of salt in the initial solution

First, we need to find the moles of the salt in the initial 0.50 M solution, using the information given. We are given the mass of the salt (4.5 g) and the concentration of the solution (0.50 M). To find the moles of salt, we can use the formula: \[ Moles\ of\ salt = \frac{mass\ of\ salt}{molar\ mass\ of\ salt} \] We are not given the molar mass of the salt, but we can get that information from the concentration and volume relationship: \[ Concentration = \frac{Moles\ of\ salt}{Volume\ of\ solution} \] Let's find the volume of the solution.

Step 2: Calculate the volume of the solution

Given that the 0.50 M solution contains 4.5 g of salt, we can rearrange the concentration formula to find the volume. \[ Volume\ of\ solution = \frac{Moles\ of\ salt}{Concentration}\] Substituting the mass and concentration, \[ Volume\ of\ solution = \frac{\frac{4.5 \ g\ of\ salt}{molar\ mass\ of\ salt}}{0.50 \frac{mol}{L}}\]

Step 3: Calculate the mass of salt in the new solution

Now, we have the same volume of a 2.50 M solution, and we need to find the mass of the salt in it. We can use the rearranged concentration formula again: \[ Moles\ of\ salt = Concentration \times Volume\ of\ solution \] \[ Moles\ of\ salt = 2.50 \frac{mol}{L} \times \frac{4.5 \ g}{0.50\ molar\ mass\ of\ salt} \] Now, we can find the mass of the salt in the new solution by multiplying the new moles of salt with the molar mass of the salt. \[ Mass\ of\ salt = Moles\ of\ salt \times molar\ mass\ of\ salt \] \[ Mass\ of\ salt = 2.50 \times \frac{4.5 \ g}{0.50} \]

Step 4: Calculate the value

Finally, performing the calculation, \[ Mass\ of\ salt = 2.50 \times 9\] \[ Mass\ of\ salt = 22.5\ g\] The mass of the salt in the same volume of a 2.50 M solution is 22.5 g.

Key Concepts

Molarity

Molarity is an essential concept in chemistry that measures the concentration of a solution. It is defined as the number of moles of solute per liter of solution. The formula to determine molarity (\(M\)) is \[M = \frac{n}{V}\]where:

• \(n\) is the number of moles of solute, and
• \(V\) is the volume of the solution in liters.

For instance, preparing a \(500 \text{ mL}\) solution with a molarity of \(0.10 \text{ M}\) means the solution has a concentration of \(0.10 moles/liter\). If you accidentally spill some, it’s critical to note that the actual molarity is unchanged, provided no solute is lost. The terms "molarity" and "concentration" are frequently interchangeable in introductory chemistry courses. Keep in mind that in any solution, if you only lose solvent (the liquid component), the concentration of your solute will remain consistent.

Evaporation Effect

Evaporation is a natural process where liquid molecules escape into the gas phase. When this happens in a solution, particularly in chemistry labs, it plays a significant role in increasing the concentration of the solution left behind. This is because while the solvent (often water) evaporates, the solute (such as salt) remains, leading to a smaller volume for the same amount of solute.

• The general effect is that the concentration (molarity) becomes higher because you have the same number of moles in a reduced amount of solvent. For example, starting with a \(500 \text{ mL}\) solution of \(0.10 \text{ M}\), and if water evaporates to leave you with \(250 \text{ mL}\), the concentration nearly doubles since the amount of salt is unchanged.

The evaporation effect is crucial in managing solutions, especially if precision in concentration is needed. It highlights the importance of covering solutions when precision is vital to prevent involuntary changes due to solvent loss.

Volume-Volume Relationship

Volume plays a pivotal role in calculating and understanding solution concentration. When dealing with chemical solutions, knowing how to manipulate volumes while assessing concentrations is valuable. A good example of this is transitioning between different concentrations and computing the mass of solutes such as in the problem provided:

• For instance, given a \(0.50 \text{ M}\) solution with \(4.5 \text{ g}\) of salt, if it’s known that this solution's volume remains constant but transitions into a \(2.50 \text{ M}\) concentration, one way to determine the new mass of the solute utilizes ratios of concentrations.
• The relationship is expressed via the rearranged equation \(\text{Mass of salt} = \text{Moles of salt} \times \text{Molar mass of salt}\), where molarity increases simultaneously with solute mass depending on the steady volume offset.

These relationships are fundamental in understanding how volume suffices and alterations regarding solute concentration dictate the solution behavior. Consequently, it enables precise calculations and adjustments for chemical experimentations and applications.