Question

A 414-mL sample of \(0.196 M \operatorname{MgBr}_{2}\) solution is prepared in a large flask. A 43.0 -mL portion of the solution is then placed into an empty \(100.0-\mathrm{mL}\) beaker. What is the concentration of the solution in the heaker?

Short answer

The concentration remains 0.196 M in the beaker.

Step-by-step solution

Understand the Initial Concentration

The initial concentration of the solution is given as \(0.196\, M\). This means there are \(0.196\) moles of \(\operatorname{MgBr}_{2}\) per liter of solution. Since we are starting with this fixed concentration, this is crucial for determining the concentration after transferring to the beaker.

Define the Amount Transferred

We have a \(43.0\, \text{mL}\) portion of the \(0.196\, M\) solution being transferred into a \(100.0\, \text{mL}\) beaker. Since this portion is being taken directly from the original solution, its concentration remains the same until any further dilution or mixing.

Calculate the Concentration in the Beaker

When the solution is transferred without adding additional solvent, the concentration remains unchanged. Therefore, the concentration of the \(\operatorname{MgBr}_{2}\) solution in the beaker is still \(0.196\, M\). This is because no dilution or chemical reaction occurs to alter the concentration.

Key concepts

Molarity

Molarity is a way to express the concentration of a solution, describing the amount of solute in a given volume of solution. It is typically expressed in moles of solute per liter of solution, denoted by the symbol "M". In simpler terms, molarity tells us how many moles of a solute are present in one liter of the solution.
To calculate molarity, use the formula:

• \(M = \frac{n}{V}\)

where:

• \(M\) is the molarity
• \(n\) is the number of moles of solute
• \(V\) is the volume of solution in liters

Understanding molarity is key when dealing with solutions, as it helps you analyze how concentrated a solution is.Moreover, molarity can also be used to determine how reactions will proceed, affecting reaction rates and product formation.

Dilution

Dilution involves adding more solvent to a solution, which results in a lower concentration of solute. An important thing to remember about dilution is that the amount of solute remains constant; only the volume of the solution changes.
When calculating dilutions, you can use the formula:

• \(M_{1}V_{1} = M_{2}V_{2}\)

where:

• \(M_{1}\) and \(V_{1}\) are the molarity and volume of the original solution
• \(M_{2}\) and \(V_{2}\) are the molarity and volume after dilution

In the original exercise example, we didn't actually perform a dilution because the solution was simply transferred from one container to another without adding additional solvent. The concentration did not change as a result.

Concentration Calculation

Calculating concentration involves determining how much solute is present within a given volume of solution. In our context, this means finding out the molarity of a solution or the amount of solute in moles per liter.
In cases where the solution's volume changes but the solute amount stays the same, dilution formulas come into play. However, as discussed in the origin exercise, since the solution was just transferred without any change in volume or solvent addition, the concentration stayed at \(0.196 \,M\).
Calculating concentration is essential in experiments and practical applications as it guides you in preparing solutions with specific properties. This ensures reactions occur as expected and safety regulations are maintained.

Chemistry Problem Solving

Problem solving in chemistry often requires a step-by-step approach to understand and solve complex problems.
Here are some key steps generally involved:

• Identify what is given and what needs to be found.
• Utilize the relevant formulas (like molarity or dilution) related to the problem.
• Carefully substitute the given values into these equations.
• Solve for the unknown variable.

In the provided exercise, since we confirmed no dilution occurred, understanding the initial molarity and recognizing that the transfer process doesn't inherently alter concentration were critical steps. Developing these problem-solving skills can greatly aid in tackling various chemistry challenges, ensuring you're prepared for both academia and real-world applications.