Chapter 8: Problem 30
If 3.5 L of a 4.8 M SrCl2 solution is diluted to 45 L, what is the molarity of the diluted solution?
Short Answer
Expert verified
The molarity of the diluted solution is 0.3733 M.
Step by step solution
01
Understanding the Concept of Dilution
Dilution is the process of making a solution less concentrated. The number of moles of solute remains constant before and after dilution. This is characterized by the formula: \( M_1V_1 = M_2V_2 \), where \(M_1\) and \(V_1\) are the molarity and volume of the initial solution, and \(M_2\) and \(V_2\) are the molarity and volume of the final solution.
02
Identify Initial Molarity and Volume
From the problem, the initial molarity, \(M_1\), is 4.8 M and the initial volume, \(V_1\), is 3.5 L.
03
Identify Final Volume
The diluted solution's final volume, \(V_2\), is 45 L.
04
Apply the Dilution Formula
Use the dilution formula \( M_1V_1 = M_2V_2 \) to solve for the final molarity, \(M_2\), by substituting the known values: \( (4.8 \text{ M}) * (3.5 \text{ L}) = M_2 * (45 \text{ L}) \).
05
Calculate the Final Molarity
Solve for \(M_2\) by dividing both sides of the equation by 45 L: \( M_2 = \frac{(4.8 \text{ M}) * (3.5 \text{ L})}{45 \text{ L}} \).
06
Final Calculation
Performing the calculation yields: \( M_2 = \frac{(16.8 \text{ mol})}{45 \text{ L}} = 0.3733 \text{ M} \). Therefore, the molarity of the diluted solution is 0.3733 M.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molarity
Molarity, often represented by the letter 'M', is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. Understanding molarity is crucial because it allows chemists to predict the number of reacting particles in a given volume, which is essential in stoichiometry. The formula to calculate molarity (\text{M}) is:
\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \].
It’s important to ensure that the volume of the solution is in liters when calculating molarity to maintain consistency in units. For our example, we consider the initial molarity of the SrCl2 solution to be 4.8 M, meaning that there are 4.8 moles of SrCl2 for every liter of solution prior to dilution.
\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \].
It’s important to ensure that the volume of the solution is in liters when calculating molarity to maintain consistency in units. For our example, we consider the initial molarity of the SrCl2 solution to be 4.8 M, meaning that there are 4.8 moles of SrCl2 for every liter of solution prior to dilution.
Dilution Formula
When a solution is diluted, its volume increases, but the amount of solute does not change. Hence, molarity will decrease. The dilution formula is an essential tool in chemistry to determine the new concentration of a solution after it has been diluted. The formula is:
\[ M_1V_1 = M_2V_2 \],
where \(M_1\) is the initial molarity, \(V_1\) is the initial volume, \(M_2\) is the final molarity, and \(V_2\) is the final volume. By rearranging the formula, we can solve for any of the variables, allowing us to find the unknown concentration or volume. In the example, we used the dilution formula to calculate the molarity of the solution after increasing its volume from 3.5 L to 45 L.
\[ M_1V_1 = M_2V_2 \],
where \(M_1\) is the initial molarity, \(V_1\) is the initial volume, \(M_2\) is the final molarity, and \(V_2\) is the final volume. By rearranging the formula, we can solve for any of the variables, allowing us to find the unknown concentration or volume. In the example, we used the dilution formula to calculate the molarity of the solution after increasing its volume from 3.5 L to 45 L.
Molality
Molality, represented by 'm', is another measure of concentration of solutions. Unlike molarity, molality measures the number of moles of solute per kilogram of solvent. It's given by the formula:
\[ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \].
Molality is particularly useful when dealing with temperature changes because, unlike volume, mass does not change with temperature. However, our exercise revolves around molarity and not molality, so we focus on volume-based concentration. In practical scenarios, both molarity and molality can be important depending on the experimental conditions.
\[ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \].
Molality is particularly useful when dealing with temperature changes because, unlike volume, mass does not change with temperature. However, our exercise revolves around molarity and not molality, so we focus on volume-based concentration. In practical scenarios, both molarity and molality can be important depending on the experimental conditions.
Concentration of Solutions
The concentration of a solution is a broad term that describes the amount of solute present in a given quantity of solvent or solution. It can be expressed in various ways, including molarity and molality as mentioned, or even in mass percentages, parts per million (ppm), or grams per liter. Understanding the concentration is vital as it helps in preparing solutions with precise properties, necessary for reactions and product formation in various industrial and laboratory processes. Concentration adjustments, such as diluting the solutions, are routinely done to reach desired levels for different applications. For instance, in the textbook exercise, we adjusted the concentration of an SrCl2 solution from 4.8 M to determine its new concentration after dilution to 45 L.