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 remains unchanged. (b) The concentration of the solution left in the container increases. (c) The mass of the salt present in the same volume of a 2.50 M solution is 22.5 grams.

Step 1: Part (a)

In this scenario, some of the 0.10 M solution of the salt is spilled. As a result, the volume of the solution decreases. However, when the solution is spilled, both the solute and the solvent are spilled, and therefore the moles of solute relative to the new volume remain constant. In other words, the concentration of the solution left in the container remains unchanged.

Step 2: Part (b)

In this case, we have a 0.10 M aqueous solution of salt, and some water evaporates over time. As water evaporates, the volume of the solvent decreases, but the amount of solute (salt) remains constant. The moles of solute (n) do not change, but the volume (V) decreases. Consequently, using the formula \( C=\frac{n}{V} \), we can conclude that the concentration of the solution left in the container increases.

Step 3: Part (c) - Calculate initial moles of solute

First, we need to determine the moles (n) of the salt present in the initial solution (0.50 M) with a mass of 4.5 grams. In order to do that, we can use the equation: \( C = \frac{n}{V} \) Since we want to find the moles of solute in the same volume of both solutions, we can rearrange the equation to solve for n: \[ n_{initial} = CV_{initial} \Longrightarrow n_{initial} = 0.50 M * V_{initial} \] Now we need to find out the mass-to-mole ratio of the salt in order to get the number of moles of salt in the initial solution.

Step 4: Part (c) - Calculate mass-to-mole ratio

We will use the mass of salt (4.5 g) in the 0.50 M solution to calculate the mass-to-mole ratio for the same volume: \[ mass\_to\_mole\_ratio = \frac{4.5 g}{n_{initial}} \]

Step 5: Part (c) - Calculate mass in the 2.50 M solution

We know the number of moles of solute in the same volume of a 2.50 M solution (n_final) can be calculated by: \[ n_{final} = CV_{final} \Longrightarrow n_{final} = 2.50 M * V_{final} \] Since V_initial = V_final, we can relate \(n_{initial}\) and \(n_{final}\) with the equation: \( \frac{n_{final}}{n_{initial}} = \frac{2.50 M}{0.50 M} \) Solving for \(n_{final}\) we get: \( n_{final} = 5 n_{initial} \) Using the mass-to-mole ratio, we can solve for the mass of salt in the 2.50 M solution (m_final): \[ m_{final} = mass\_to\_mole\_ratio * n_{final} = \frac{4.5 g}{n_{initial}} * 5 n_{initial} \] Cancelling out \(n_{initial}\) and simplifying, we get: \[ m_{final} = 22.5 g \] Thus, the mass of the salt present in the same volume of a 2.50 M solution is 22.5 grams.

Key Concepts

Solution concentration

Solution concentration is a measure of how much solute is dissolved in a specific amount of solvent or solution. It is expressed as molarity (M), which is the number of moles of solute per liter of solution. The formula for calculating molarity is:

• \( M = \frac{n}{V} \)

where \( n \) is the number of moles of solute and \( V \) is the volume of the solution in liters.
When preparing a solution, we must know both the amount of solute and the volume of the solvent to achieve a desired molarity. For example, preparing a 0.10 M solution requires dissolving a specific amount of solute in 1 liter of solvent to achieve that concentration.
Solution concentration is crucial in chemical reactions and laboratory processes. It helps in determining the right proportions of substances involved in reactions and ensures consistent results. Maintaining a specific concentration is necessary to get the expected outcomes in experiments.

Evaporation effects on concentration

Evaporation is a process in which a liquid turns into vapor, and it has a significant impact on the concentration of a solution. When a solution is left uncovered, the solvent (usually water in aqueous solutions) can evaporate.
While the solvent decreases due to evaporation, the amount of solute remains the same. This leads to an increase in the concentration of the solution because the same amount of solute is now present in a smaller volume. The mathematical representation is:

• \( C = \frac{n}{V} \)

where \( n \) is unchanged, but \( V \) decreases, causing \( C \) (concentration) to increase.
This principle is important in practical applications where solvents can naturally evaporate over time. In such cases, the concentration of the solution will become stronger unless adjustments are made to bring it back to the desired level by adding more solvent.

Moles of solute calculation

Calculating the moles of solute is fundamental when dealing with solutions of different concentrations. Moles are a measure of the number of particles in a substance. To find out how many moles are present in a solution, we use the equation:

• \( n = M \times V \)

where \( n \) is the number of moles, \( M \) is the molarity, and \( V \) is the volume of the solution in liters.
In cases where the concentration changes, it is possible to determine how the moles relate by comparing the initial and final conditions. In the exercise provided:
a 0.50 M solution with a known mass of solute (4.5 g) had its moles derived. This was then used to find the mass in a 2.50 M solution of the same volume through the relation:

• \( \frac{n_{final}}{n_{initial}} = \frac{2.50 \, M}{0.50 \, M} \)

allowing for the calculation of a new mass based on increased concentration, demonstrating how moles can be utilized to track changes in solution properties.