Question

A \(60.0-\mathrm{mL} 0.513 \mathrm{M}\) glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) solution is mixed with \(120.0 \mathrm{~mL}\) of \(2.33 \mathrm{M}\) glucose solution. What is the concentration of the final solution? Assume the volumes are additive.

Short answer

The concentration of the glucose solution after mixing is 1.724 M.

Step-by-step solution

Calculate Moles in Each Solution

The concentration of a solution tells us how many moles of solute are present in every liter of solution. By multiplying the volume of solution (converting it from milliliters to liters) by the concentration, the number of moles of glucose in each solution can be calculated. For the 60.0 mL of 0.513 M glucose solution: moles = 0.060 L * 0.513 mol/L = 0.03078 mol and For the 120.0 mL of 2.33 M glucose solution: moles = 0.120 L * 2.33 mol/L = 0.2796 mol

Calculate Total Moles and Total Volume

The total number of moles of glucose is the sum of the moles in each solution (0.03078 mol + 0.2796 mol = 0.31038 mol). The total volume of the mixture is also the sum of the volumes of each solution (60 ml + 120 ml = 180 ml = 0.18 L).

Calculate Final Concentration

The concentration of the final solution can now be calculated by dividing the total number of moles of glucose by the total volume of the mixture in liters: concentration = moles/volume = 0.31038 mol / 0.18 L = 1.724 M

Key concepts

Understanding Molarity

Molarity is a fundamental concept in chemistry that helps us understand how concentrated a solution is. It tells us how many moles of a solute—such as glucose in this case—are present in one liter of solution. To think about it simply, it's like saying how much sugar is dissolved in a fixed amount of water.

Here's a quick refresher on what a mole is: it's a unit that chemists use to count very small things like atoms and molecules. When we refer to a concentration of 0.513 M (molarity), we mean there are 0.513 moles of glucose in one liter of that solution.

This concept is incredibly useful because it allows scientists to predict how solutions will behave during reactions. Understanding molarity empowers you to determine how much of a substance is needed to achieve a desired reaction outcome. And remember, practice makes perfect. Try playing around with different concentrations in practice problems to see how they change the final molarity of mixed solutions.

Steps for Moles Calculation

Calculating the number of moles is an essential skill in chemistry. It gives us the ability to precisely articulate how much of a material is present in a given mixture. This process becomes particularly crucial when mixing substances with different concentrations.

To find the number of moles, use the equation: \(\text{moles} = \text{volume in liters} \times \text{molarity}\ \). Let's break this down with glucose solution as an example:

• The 60.0 mL glucose solution: Convert 60 mL to liters, which becomes 0.060 L. The moles are calculated as \(0.060 \text{ L} \times 0.513 \text{ mol/L} = 0.03078 \text{ mol}\).
• The 120.0 mL glucose solution: Convert 120 mL to liters, which is 0.120 L. The moles calculation is \(0.120 \text{ L} \times 2.33 \text{ mol/L} = 0.2796 \text{ mol}\).

When you sum these moles, 0.03078 mol + 0.2796 mol = 0.31038 mol of glucose is present in the combined solution.

Converting Volumes for Accurate Calculations

Successful chemistry problem-solving often hinges on proper conversions, especially when dealing with volumes. In many cases, you'll start with a volume expressed in milliliters (mL), but calculations typically require the volume to be in liters (L), since molarity uses liters as the base unit.

To convert milliliters to liters, simply divide by 1000. For example, for the mixture:

• 60.0 mL becomes 0.060 L = 60.0 mL / 1000
• 120.0 mL converts to 0.120 L = 120.0 mL / 1000

After calculating the total volume of the solution by adding the converted volumes together (0.060 L + 0.120 L = 0.180 L), you can use this as the volume in molarity calculations.

Being diligent with your volume conversions ensures your calculations make sense and are correctly aligned with the units used in scientific equations. This is a simple yet vital step in ensuring the accuracy of your results.