A solution of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) in water is prepared by dissolving \(75.0 \mathrm{~mL}\) of ethanol (density \(=0.79 \mathrm{~g} / \mathrm{cm}^{3}\) ) in enough water to make \(250.0 \mathrm{~mL}\) of solution. What is the molarity of the ethanol in this solution?

Short Answer

Expert verified
The molarity of ethanol in the solution can be calculated as \(\text{Molarity of ethanol} = \frac{75.0 \times 0.79}{46.07 \times 0.250} \mathrm{M}\). After performing the calculations, we find the final molarity of ethanol in the solution.

Step by step solution

01

Calculate the mass of ethanol

First, we'll calculate the mass of ethanol using its volume and density. Use the given volume (75.0 mL) and density (0.79 g/cm³): \[ \text{Mass of ethanol} = \text{volume} \times \text{density} = 75.0 \text{ mL} \times 0.79\frac{\text{g}}{\text{cm}^3} \] Since \(1 \text{ cm}^3 = 1 \text{ mL}\), we can convert the density to g/mL and perform the calculation: \[ \text{Mass of ethanol} = 75.0 \text{ mL} \times 0.79\frac{\text{g}}{\text{mL}} \]
02

Determine the number of moles of ethanol

Next, we'll determine the number of moles of ethanol by dividing the mass of ethanol by its molar mass. Ethanol's molar mass is: \[ \text{C}_{2}\text{H}_{5}\text{OH} = 2 \times 12.01 + 6 \times 1.01 + 16.00 = 46.07 \text{ g/mol} \] Now, divide the mass of ethanol by the molar mass to find the number of moles: \[ \text{Moles of ethanol} = \frac{\text{mass of ethanol}}{\text{molar mass}} = \frac{75.0 \times 0.79 \text{ g}}{46.07 \text{ g/mol}} \]
03

Calculate the total volume of the solution

To find the total volume of the solution, convert the given volume in milliliters (250.0 mL) to liters: \[ \text{Total volume of the solution} = 250.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.250 \text{ L} \]
04

Calculate the Molarity of the solution

Finally, we can calculate the molarity of ethanol in the solution by dividing the moles of ethanol by the total volume of the solution: \[ \text{Molarity of ethanol} = \frac{\text{moles of ethanol}}{\text{total volume of the solution}} \] Substitute the values we found in the previous steps: \[ \text{Molarity of ethanol} = \frac{75.0 \times 0.79}{46.07 \times 0.250} \mathrm{M} \] Now, perform the calculations to find the final molarity of ethanol in the solution.

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Key Concepts

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

Solution Concentration
Understanding solution concentration is essential in chemistry, as it describes the amount of a substance, or solute, present in a given volume of solvent. The concentration can be expressed in various ways, including molarity, molality, normality, and percentage composition, among others. Molarity, which is our focus, is a common unit of concentration used to express the number of moles of solute per liter of solution (mol/L). It's a crucial concept because it allows chemists to measure reaction rates, determine equilibrium constants, and perform titration calculations.

To grasp the concept of molarity, let's consider a simple analogy. Imagine we have a lemonade stand where the 'lemon syrup' is our solute and the water is our solvent. If we mix a certain amount of lemon syrup with a fixed amount of water, we can describe the concentration of the lemon syrup in terms of 'lemon syrup per jug of water'. Translating this to chemistry, if we dissolve a specific amount of a chemical in a certain volume of water, we can describe its concentration in moles per liter.
Molar Mass
Molar mass is an indispensable value in chemistry, representing the mass of one mole of a substance. It's expressed in grams per mole (g/mol) and is calculated by summing the atomic masses of each element in a compound. Each element's atomic mass can be found on the periodic table and must be multiplied by the number of atoms of that element in the formula.

For instance, ethanol (C2H5OH) has a molar mass of 46.07 g/mol. Understanding molar mass is key to converting between the mass of a substance and the number of moles, enabling much of the quantitative analysis in chemistry such as stoichiometry. Think of it as a conversion factor that links the microscopic world of atoms and molecules to weights and measures that we can manipulate in the lab.
Density to Mass Conversion
In chemistry, density is a measure of mass per unit of volume and often has the units grams per milliliter (g/mL) or grams per cubic centimeter (g/cm³). Density to mass conversion is a practical skill, especially when dealing with solutions. If you know the volume of a liquid and its density, you can easily calculate the mass.

To perform such a conversion, simply multiply the volume of the liquid by its density. This process is akin to determining how much the liquid 'weighs'. For our ethanol example, with a volume of 75.0 mL and a density of 0.79 g/mL, the mass of ethanol can be found quickly. Always remember that 1 mL is equivalent to 1 cm³, so density in g/mL is numerically equivalent to g/cm³, reassuring us that the units we use won't change the outcome of the calculation.
Moles of Solute
The mole is a fundamental concept in chemistry, serving as a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure. A mole represents Avogadro's number (approximately 6.022 x 10²³) of particles, whether they are atoms, ions, or molecules. When we talk about 'moles of solute', we are referring to the amount of that substance present in a given solution.

Calculating moles of solute is straightforward once you have the mass and the molar mass: simply divide the former by the latter. It's a process similar to determining how many dozen (12) items you have if you know the total number. If you have 24 apples, knowing that a 'dozen' equals 12 helps you recognize that you have 2 dozen apples. Similarly, if you know you have a certain grams of a chemical, knowing the molar mass (grams per mole) helps you determine the number of moles you have.
Chemistry Education
The role of chemistry education is pivotal in developing future scientists and informed citizens. It goes beyond rote memorization of facts and formulae; indeed, it's about cultivating a deep understanding of the materials and processes that affect our everyday lives. Chemistry education enables learners to grasp crucial concepts such as solution concentration, molar mass, and the quantification of reactions.

Effective chemistry education emphasizes hands-on learning, critical thinking, and real-world applications. To aid in this, it's beneficial to relate abstract concepts to concrete examples, much like how the earlier analogy of lemonade helped to clarify the notion of solution concentration. It's also about providing clear, step-by-step explanations of problem-solving techniques, as we've done here with our dissolution and molarity calculation. Encouraging curiosity and connecting chemistry to everyday experiences can make learning both engaging and illuminating.

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Most popular questions from this chapter

Douglasite is a mineral with the formula \(2 \mathrm{KCl} \cdot \mathrm{FeCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\). Calculate the mass percent of douglasite in a \(455.0-\mathrm{mg}\) sample if it took \(37.20 \mathrm{~mL}\) of a \(0.1000 \mathrm{M} \mathrm{AgNO}_{3}\) solution to precipitate all the \(\mathrm{Cl}^{-}\) as \(\mathrm{AgCl}\). Assume the douglasite is the only source of chloride ion.

A mixture contains only \(\mathrm{NaCl}\) and \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3} .\) A \(0.456-\mathrm{g}\) sample of the mixture is dissolved in water, and an excess of \(\mathrm{NaOH}\) is added. producing a precipitate of \(\mathrm{Fe}(\mathrm{OH})_{3} .\) The precipitate is filtered. dried, and weighed. Its mass is \(0.107 \mathrm{~g}\). Calculate the following. a. the mass of iron in the sample b. the mass of \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) in the sample c. the mass percent of \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) in the sample.,

Saccharin \(\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S}\right)\) is sometimes dispensed in tablet form. Ten tablets with a total mass of \(0.5894 \mathrm{~g}\) were dissolved in water. The saccharin was oxidized to convert all the sulfur to sulfate ion, which was precipitated by adding an excess of barium chloride solution. The mass of \(\mathrm{BaSO}_{4}\) obtained was \(0.5032 \mathrm{~g}\). What is the average mass of saccharin per tablet? What is the average mass percent of saccharin in the tablets?

The drawings below represent aqueous solutions. Solution A is \(2.00 \mathrm{~L}\) of a \(2.00 \mathrm{M}\) aqueous solution of copper(II) nitrate. Solution \(\mathrm{B}\) is \(2.00 \mathrm{~L}\) of a \(3.00 \mathrm{M}\) aqueous solution of potassium hydroxide. a. Draw a picture of the solution made by mixing solutions \(\mathrm{A}\) and \(\mathrm{B}\) together after the precipitation reaction takes place. Make sure this picture shows the correct relative volume compared to solutions \(\mathrm{A}\) and \(\mathrm{B}\), and the correct relative number of ions, along with the correct relative amount of solid formed. b. Determine the concentrations (in \(M\) ) of all ions left in solution (from part a) and the mass of solid formed.

A 6.50-g sample of a diprotic acid requires \(137.5 \mathrm{~mL}\) of a \(0.750\) \(M\) NaOH solution for complete neutralization. Determine the molar mass of the acid.

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