Question

A standard solution is prepared for the analysis of fluoxymesterone \(\left(\mathrm{C}_{20} \mathrm{H}_{29} \mathrm{FO}_{3}\right),\) an anabolic steroid. A stock solution is first prepared by dissolving \(10.0 \mathrm{mg}\) of fluoxymesterone in enough water to give a total volume of \(500.0 \mathrm{mL}\). A \(100.0-\mu \mathrm{L}\) aliquot (portion) of this solution is diluted to a final volume of \(100.0 \mathrm{mL} .\) Calculate the concentration of the final solution in terms of molarity.

Short answer

The concentration of the final solution in terms of molarity is \(5.944 \times 10^{-7} \mathrm{M}\).

Step-by-step solution

Calculate the moles of fluoxymesterone

To calculate the moles of fluoxymesterone, we will use the molecular formula provided and the given mass of fluoxymesterone: Molecular formula: C_20H_29FO_3 Molecular weight = (20 × 12.01) + (29 × 1.01) + 1 × 19 + 3 × 16 = 240.2 + 29.29 + 19 + 48 = 336.49 g/mol Given mass = 10.0 mg = 0.0100 g Using the formula: moles = mass / molecular weight moles of fluoxymesterone = 0.0100 g / 336.49 g/mol = 2.972 × 10 ^ (-5) mol

Calculate the concentration of the initial solution

Now, we will calculate the concentration of fluoxymesterone in the initial solution using the formula: Molarity (M) = moles of solute / volume of solution Given volume of initial solution = 500.0 mL = 0.5000 L Molarity = 2.972 × 10 ^ (-5) mol / 0.5000 L = 5.944 × 10 ^ (-5) M

Calculate the molar concentration of the final solution

To calculate the molar concentration of the final solution, we will use the dilution formula: M1 × V1 = M2 × V2 Where: M1 = molar concentration of initial solution (5.944 × 10 ^ (-5) M) V1 = volume of aliquot (100.0 µL = 0.000100 L) M2 = molar concentration of final solution (unknown) V2 = volume of final solution (100.0 mL = 0.1000 L) Rearrange the formula to solve for M2: M2 = (M1 × V1) / V2 M2 = (5.944 × 10 ^ (-5) M) × (0.000100 L) / 0.1000 L M2 = 5.944 × 10 ^ (-7) M Therefore, the concentration of the final solution in terms of molarity is 5.944 × 10 ^ (-7) M.

Key concepts

Standard Solution Preparation

Preparing a standard solution involves accurately measuring and combining a known quantity of a solute with a specific volume of solvent to achieve a desired molarity. The process is critical in analytical chemistry for creating solutions with precise concentrations, which are used for various tests and calibrations.

In the exercise, a stock solution is made by dissolving fluoxymesterone, an anabolic steroid, to achieve a known concentration. For accurate results, all measurements must be precise. The fluoxymesterone is weighed to the nearest milligram and dissolved in a volumetric flask filled to a defined mark to ensure the total volume is exactly 500.0 mL. Accuracies in these preparatory steps are essential to ensure a reliable standard solution for subsequent dilutions and analyses.

The Mole Concept

The mole concept is a fundamental principle in chemistry that provides a bridge between the atomic world and the macroscopic world we can measure. One mole of any substance contains Avogadro's number of particles - approximately 6.022 x 10²³ entities, whether they are atoms, molecules, ions, or electrons.

In this exercise, we calculate the number of moles of fluoxymesterone by dividing the mass of the substance by its molecular weight. The molecular weight is the sum of the atomic masses of all the atoms in the molecule. Understanding how to use the mole concept allows us to relate the mass of the substance to the number of its particles, which is crucial for the preparation of solutions with desired molar concentrations.

Dilution of Solutions

Dilution is the process of reducing the concentration of a solute in a solution, usually by adding more solvent. The core principle is that the amount of solute remains constant before and after the dilution. As such, dilutions do not affect the actual number of moles of solute, just the concentration.

Using the formula \( M1 \times V1 = M2 \times V2 \), where \( M1 \) and \( M2 \) are the concentrations before and after dilution, and \( V1 \) and \( V2 \) are the volumes before and after dilution, we can calculate the new concentration after dilution, as demonstrated in the exercise. Dilution calculations are common in preparing solutions for laboratory experiments where precise concentrations are required.