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) \(\mathrm{A}\) 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?

Short answer

In summary: (a) The concentration of the solution left in the container remains the same at \(0.10 M\) after spilling some of it. (b) The concentration of the solution increases when water evaporates from the uncovered solution. (c) In the new \(2.50 M\) solution, the mass of the salt present in the same volume is \(9.0 g\).

Step-by-step solution

Define the concentration

The concentration of a solution is defined as the ratio of the amount of solute to the volume of the solution. In this case, the concentration is given as \(0.10 M\), which indicates the moles of solute present per liter of the solution.

Determine the effect of spilling on the concentration

When you spill some of the solution, the amount of solute remains unchanged while the volume of the solution decreases. The amount of solute is not affected by the spill, so the concentration of the solution left in the container stays the same at \(0.10 M\). #b. Analyzing the concentration after evaporation of the solution#

Define the effect of evaporation on the solution

When a solution is left uncovered, and water evaporates, the volume of the solution decreases. The amount of solute in the solution remains the same, as evaporation only affects the solvent (water, in this case).

Determine the effect of evaporation on the concentration

As water evaporates and the volume of the solution decreases while the amount of solute remains constant, the concentration of the solution will increase. #c. Finding the mass of solute in a new solution#

Calculate the moles of solute in the initial solution

According to the problem, we have a \(0.50 M\) solution and it contains \(4.5 g\) of a salt in a certain volume. We can determine the number of moles of the solute as follows: Moles of solute = \(\frac{4.5 g}{Molar\,mass\,of\,solute}\)

Calculate the moles of solute in the new solution

The volume of the new solution is the same as the volume of the old solution, and the concentration has increased to \(2.50 M\). The moles of solute in the new solution can be calculated as the product of its concentration and the volume: Moles of solute in new solution = \(2.50 M \times Volume\)

Determine the mass of solute in the new solution

Since we have the moles of solute in the new solution, we can find the mass of the solute by multiplying the moles by the molar mass of the solute: Mass of solute in new solution = \(Moles\,of\,solute\,in\,new\,solution \times Molar\,mass\,of\,solute\) But we have a relationship between the moles of solute in the old and new solutions. The amount of solute in the old solution (\(0.50 M\)) is half of the amount of solute in the new solution (\(2.50 M\)). Hence, Mass of solute in new solution = \(2 \times 4.5 g = 9.0 g\)

Key concepts

Molarity

Understanding molarity is critical for chemists and students alike as it describes the concentration of a solute in a solution. In its essence, molarity (M) measures how many moles of a substance are present in one liter of solution. It's the backbone of concentration calculations in chemistry. For instance, a solution with a molarity of 0.10 M contains 0.10 moles of the solute in every liter.

In practical terms, if a student prepares 500 mL of a 0.10 M salt solution, they effectively have 0.05 moles of the salt in the mixture – found by multiplying the volume in liters (0.5 L) by the molarity (0.10 M). Molarity provides a straightforward way to understand the ratio between the quantity of the dissolved substance and the total volume of the solution, which is vital when analyzing chemical reactions and processes.

Evaporation and Concentration

Evaporation plays an intriguing role in the science of solutions. When a liquid solvent like water evaporates from a solution, the remaining solution becomes more concentrated. The key point to grasp here is that evaporation only affects the solvent, not the solute.

Picture a solution that's left open. As water particles escape into the air, the total volume of the solution decreases, but the amount of solute stays the same. This change results in an increased molarity. For a student observing their 0.10 M salt solution become more concentrated after evaporation, it means the molarity is higher than 0.10 M after water leaves the beaker. The original solution setup allows students to witness this transformation firsthand, emphasizing the impact of evaporation on concentration without adding or removing the solute itself.

Solute and Solvent

The duo of solute and solvent defines any solution. A solute is a substance that is dissolved in another substance, while a solvent is the medium in which the solute is dissolved. In many cases, like with saltwater, the solute is a solid (salt) and the solvent is a liquid (water).

When a solute dissolves in a solvent, it breaks into particles and becomes evenly distributed throughout the solvent, forming a homogeneous mixture. The ratio of solute to solvent determines the solution's concentration. This ratio is pivotal because it influences the properties of the solution such as boiling point, freezing point, and vapor pressure. Through the demonstration of how concentration does not change when the solution is spilled (solute stays, volume changes), students comprehend that concentration is a measure related to both solute and solvent quantities.

Concentration Calculations

Calculating the concentration of a solution is a practical skill in chemistry. When tasked with determining the mass of a solute in a solution from its molarity, it's a two-step process. First, calculate the number of moles of solute using the molarity and volume. Then, convert moles to grams using the molar mass of the solute. The exercise given showcases these calculations by comparing two solutions with different molarities but the same volume.

To understand this fully, consider a 0.50 M solution containing 4.5 g of a salt. If we have the same volume of a 2.50 M solution, it would inherently contain five times the moles of solute because the molarity is five times greater. As a result, we can compute the mass of the solute in this new solution by multiplying the initial mass by five, assuming the molar mass remains constant.