To prepare a very dilute solution, it is advisable to perform successive dilutions of a single prepared reagent solution, rather than to wcigh a very small mass or to measure a very small volume of stock chemical. A solution was prepared by transferring \(0.661 \mathrm{~g}\) of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to a \(250.0\)-mL volumetric flask and adding water to the mark. A \(1.000-\mathrm{mL}\). sample of this solution. was transferred to a 500 -mL. volumetric flask and diluted to the mark with water. Then \(10.0 \mathrm{~mL}\) of the diluted solution was transferred to a 250 -mL. flask and diluted to the mark with water. (a) What is the final concentration of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{\text {, in solution? (b) What }}\) mass of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is in this final solution? (The answer to the last question gives the amount that would have had to have been weighed out if the solution had been prepared directly.)

Step by step solution

01

Calculate the molarity of the initial solution

Determine the molarity (M) of the solution after dissolving 0.661 g of \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) in the 250.0 mL volumetric flask. First, find the molar mass of \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) using the periodic table (K = 39.10 g/mol, Cr = 52.00 g/mol, O = 16.00 g/mol). The molar mass is 2(39.10) + 2(52.00) + 7(16.00) = 294.20 g/mol. Then calculate the molarity using the formula: M = mass (g) / molar mass (g/mol) / volume (L). Convert the volume from mL to L (250.0 mL = 0.2500 L). The initial molarity is: M = 0.661 g / 294.20 g/mol / 0.2500 L.

02

Calculate the molarity after the first dilution

Use the dilution formula \(M_1V_1 = M_2V_2\) to find the molarity after the first dilution, where:\(M_1\) is the initial molarity, \(V_1\) is the volume of the sample taken (1.000 mL = 0.001000 L), \(M_2\) is the molarity after the first dilution, and \(V_2\) is the final volume of the flask (500 mL = 0.500 L). Solve for \(M_2\) to find the new molarity.

03

Calculate the molarity after the second dilution

Repeat the dilution formula with the new values: \(M_2V_2 = M_3V_3\), where:\(M_2\) is the molarity from Step 2, \(V_2\) is the volume of the diluted solution taken (10.0 mL = 0.0100 L), \(M_3\) is the final molarity, and \(V_3\) is the final volume of the second dilution flask (250 mL = 0.250 L). Solve for \(M_3\) to get the final molarity.

04

Calculate the final mass of \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) in the final diluted solution

Using the final molarity (M) from Step 3, calculate the mass of \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) that would be present in the final 250 mL solution by rearranging the molarity formula to solve for mass: mass = M * molar mass * volume. Use the final molarity (M), the molar mass from Step 1, and the final volume (0.250 L).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molarity Calculation

Molarity, often denoted as M, represents the concentration of a solute within a solution. It is defined as the number of moles of solute dissolved per liter of solution. Calculating molarity is a fundamental skill in chemistry, as it allows us to describe the strength of a solution. To carry out such a calculation, the formula used is:
\[\[\begin{align*} M = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)} \times \text{volume of solution (L)}} \end{align*}\]\]
In the context of our exercise, this involves determining the molar mass of the solute, which is potassium dichromate (\( \mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} \)), converting the volume from milliliters to liters since molarity is expressed in terms of liters, and then applying the mass of the solute used in the preparation. Once you have these values, insert them into the formula to find the molarity of the initial solution.

Serial Dilution

Serial dilution is a stepwise dilution of a substance in solution. It is commonly used in laboratories to obtain a solution of the desired concentration, which is particularly important when working with concentrations that are too small to measure accurately. A serial dilution involves diluting a known concentration of a solution multiple times by the same factor each time. Each subsequent dilution uses a portion of the previous dilution as the starting point. In our exercise, a serial dilution was performed by taking a small, measurable volume of a concentrated stock solution and diluting it in a stepwise manner to reach a much lower concentration, suitable for the final application. This method conserves reagents and reduces the risk of error when measuring very small volumes.

Molar Mass Determination

Molar mass is a critical value in many chemical calculations, representing the mass of one mole of a given substance. The molar mass is expressed in grams per mole (g/mol) and is calculated by adding together the atomic masses of each element in a compound, which can be found on the periodic table. For instance, to find the molar mass of potassium dichromate (\( \mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7} \)), we sum the molar masses of all potassium (K), chromium (Cr), and oxygen (O) atoms within the molecular formula. Laboratory work often involves using the molar mass to convert between grams and moles, enabling us to perform calculations involved in preparing solutions like the one in our problem.

Dilution Formula

The dilution formula, \( M_1V_1 = M_2V_2 \), is a mathematical representation of the conservation of mass during the process of dilution. This formula tells us that the product of the initial concentration (\( M_1 \)) and volume (\( V_1 \)) of the solution must equal the product of the final concentration (\( M_2 \)) and volume (\( V_2 \)) once diluted. This equation is pivotal in calculating the new concentration of a solution after it has been diluted, as seen in our exercise. When performing a dilution, you can use the initial molarity and volume of your concentrated solution (\( M_1 \) and \( V_1 \)), along with the volume of water or solvent you are adding (\( V_2 \)), to calculate the final molarity (\( M_2 \)) of the diluted solution.