Question

To what volume should you dilute 50.0 \(\mathrm{mL}\) of a 5.00 \(\mathrm{M} \mathrm{KI}\) solution so that 25.0 \(\mathrm{mL}\) of the diluted solution contains 3.05 \(\mathrm{g}\) of \(\mathrm{KI}\) ?

Short answer

The final volume to dilute 50.0 mL of a 5.00 M KI solution is calculated to be the volume where 25.0 mL of the diluted solution contains 3.05 g of KI.

Step-by-step solution

Calculate moles of KI in 25.0 mL of diluted solution

To find out how many moles are in 3.05 g of KI, use the molar mass of KI. The molar mass of KI (potassium iodide) is approximately 166.0 g/mol. The number of moles (n) is calculated using the formula: n = mass/molar mass. Hence, n = 3.05 g / 166.0 g/mol.

Determine the concentration of KI in the diluted solution

Since we know the number of moles in 25.0 mL of the solution, we can find the concentration (molarity, M) by using the formula: M = moles/volume(L). Since the volume needs to be in liters, we convert 25.0 mL to liters by dividing by 1000.

Calculate the total moles of KI in the original 50.0 mL solution

We use the initial concentration of the KI solution to calculate the total moles in the 50.0 mL solution before dilution. Using the formula: moles = concentration * volume(L), we find the total number of moles.

Calculate the final volume required for the dilution

The moles of KI will remain the same before and after dilution. To find the total volume required for the desired concentration, we use the formula: V_final = moles / final concentration. We can use the final concentration from Step 2, which is based on the 25 mL volume containing 3.05 g of KI.

Convert final volume back to milliliters

Since the solution asked for the final volume in milliliters, we convert the volume from liters to milliliters by multiplying by 1000.

Key concepts

Molar Mass

The molar mass is a fundamental concept in chemistry that describes the weight of one mole of a substance. It is usually expressed in grams per mole (g/mol). The molar mass is the mass of 6.02 \( \times \) 10²³ particles, atoms, or molecules of a substance (Avogadro's number), which makes it invaluable for converting between the mass of a substance and the number of moles. Understanding molar mass is key to solving problems related to solution dilution calculations, as it allows us to determine how many moles of a solute are present in a given mass. For instance, the molar mass of potassium iodide (KI) is approximately 166.0 g/mol. If you have 3.05 g of KI, you can calculate the moles of KI by dividing the mass by the molar mass: \( \frac{3.05 g}{166.0 g/mol} \).

Concentration of Solution

The concentration of a solution is typically measured in molarity (M), which is calculated as moles of solute per liter of solution. It is a measure of how much solute is contained in a solution. To calculate the concentration of a solution, use the formula \( M = \frac{moles}{volume(L)} \). In the context of our example, if we find the number of moles of KI to be in a given volume, we can then determine the molarity of that solution. This step is crucial when performing a solution dilution calculation, as it allows us to understand the relationship between the quantity of solute and the total volume of the solution after dilution.

For instance, if we have determined that 25.0 mL of a solution contains a certain number of moles of KI, we need to convert the volume of solution from milliliters to liters (since molarity is moles per liter), and then we can find the molarity of the diluted solution using the formula mentioned above.

Moles Calculation

Moles calculation is essential in chemistry for quantifying the amount of substance. One mole is Avogadro's number of particles. The mole concept bridges the microscopic world of atoms and molecules to the macroscopic world we can measure. To calculate moles from a given mass, as shown in the solution, we use the formula \( n = \frac{mass}{molar mass} \).

In this particular exercise, after determining the molar mass of KI, we calculate the moles of KI in 3.05 g to understand how much substance we are starting with before dilution. These moles calculations are critical when determining the final volume needed for dilution, as they ensure that the concentration of the diluted solution is accurately derived for practical use. By ensuring students grasp the moles calculation, they can confidently approach a variety of chemical problems involving concentrations and dilutions.

Volume Conversion

Volume conversion is often required in solution dilution problems because the standard units of concentration are moles per liter. Thus, converting from milliliters (mL) to liters (L) is necessary because lab measurements are commonly made in milliliters. The conversion is straightforward: divide the volume in milliliters by 1000 to obtain the volume in liters.

For example, to convert the 25.0 mL of diluted KI solution used in our exercise to liters, you would calculate \( \frac{25.0 mL}{1000} \) to get 0.025 L. This is necessary to correctly calculate the concentration of the solution. Similarly, after finding the necessary final volume in liters for our dilution, we convert that back to milliliters to report in common laboratory units, by multiplying the volume in liters by 1000. This skill is vital for students to master, as accurate volume conversions are crucial in making precise measurements in laboratory experiments and real-world applications.