/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 A solution of ethanol \(\left(\m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A solution of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\right)\) 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 the ethanol solution is approximately 5.144 M.

Step by step solution

01

Calculate the mass of ethanol

Given the volume and density of ethanol, we can calculate the mass of ethanol by using the formula: mass = volume × density. Here, the volume of ethanol is 75.0 mL, and the density is 0.79 g/cm³. Since 1 mL is equivalent to 1 cm³, we can keep the same unit. mass_ethanol = volume_ethanol × density_ethanol mass_ethanol = 75.0 mL × 0.79 g/mL mass_ethanol = 59.25 g The mass of ethanol is 59.25 grams.
02

Calculate the number of moles of ethanol

To find the number of moles of ethanol, we will use the molar mass of ethanol, which is given by the formula: C₂H₅OH. The molar mass of ethanol is approximately 46.07 g/mol, considering the atomic masses of carbon (12.01), hydrogen (1.01) and oxygen (16.00). number_of_moles = mass_ethanol / molar_mass_ethanol number_of_moles = 59.25 g / 46.07 g/mol number_of_moles ≈ 1.286 mol The number of moles of ethanol in the solution is approximately 1.286 moles.
03

Calculate the molarity of the ethanol solution

Molarity is defined as the number of moles of solute per liter of solution. We are given the total volume of the solution, which is 250.0 mL or 0.250 L. We can now calculate the molarity of ethanol in the solution using the number of moles and the total volume: molarity_ethanol = number_of_moles / total_volume molarity_ethanol = 1.286 mol / 0.250 L molarity_ethanol ≈ 5.144 M The molarity of the ethanol solution is approximately 5.144 M.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Solution Concentration
When discussing the composition of a mixture, we often focus on the solution concentration. This is a measure of how much solute is dissolved within a given volume of solvent. In the world of chemistry, several types of concentrations are used, but one of the most common is molarity.

Molarity (\( M \)) is defined as the number of moles of solute per liter of solution, which tells us the strength of the solution. It's a straightforward way to communicate how concentrated a solution is and is crucial for predicting how substances will react with one another in a mixture. The formula is given as: \[\text{{Molarity}} = \frac{{\text{{moles of solute}}}}{{\text{{volume of solution in liters}}}}.\]
For instance, in the exercise about ethanol, molarity tells us how concentrated the ethanol is within the water solution. Calculating the molarity helps students understand and predict the behavior of substances in real-world scenarios, such as in medical solutions, industrial processes, or laboratory experiments.
Mole Concept
The mole concept is a fundamental bridge between the microscopic world of atoms and molecules and the macroscopic world of grams and liters that we can observe. A mole is a unit that represents a quantity of material. Specifically, one mole is equal to Avogadro's number of particles (usually atoms or molecules), which is approximately \(6.022 \times 10^{23}\) particles.

This concept allows us to easily communicate and make calculations regarding the amount of substance involved in chemical reactions. When we discuss molarity, we're often dealing with moles. For example, we might say a solution contains 1.286 moles of ethanol, as was calculated in the given exercise. By using the mole concept, chemists can quantify and understand the relationships in chemical reactions.
Molar Mass
Understanding molar mass is crucial when performing calculations in chemistry, especially when working with solution concentrations and reactions. Molar mass is the weight of one mole of a substance and is expressed in grams per mole (g/mol). It's calculated by summing the atomic masses of all the atoms in a molecule.

Each element's atomic mass can be found on the periodic table and represents the average mass of a single atom of that element in atomic mass units (amu). For example, the molar mass of ethanol (\(C_2H_5OH\)) is calculated using the atomic masses of carbon, hydrogen, and oxygen. This value is critical when converting mass to moles as part of molarity calculations, just as we saw in the exercise with ethanol.
Stoichiometry
The term stoichiometry is derived from two Greek words meaning 'element' and 'measure,' and it involves the quantitative relationships of the elements in chemical reactions. It's very much like a recipe for chemistry where the required amounts of reactants to produce desired products are detailed.

In terms of the calculation process, stoichiometry allows us to predict how much of a substance will be needed or produced in a chemical reaction. It involves using balanced chemical equations to find the balance between reactants and products. The stoichiometry process involves understanding the mole concept and molar mass to ensure that atoms are conserved in the reactions and to calculate quantities such as molarity of solutions, another key point demonstrated in the ethanol solution exercise.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write net ionic equations for the reaction, if any, that occurs when aqueous solutions of the following are mixed. a. chromium(III) chloride and sodium hydroxide b. silver nitrate and ammonium carbonate c. copper(II) sulfate and mercury(1) nitrate d. strontium nitrate and potassium iodide

Consider the reaction of \(19.0 \mathrm{~g}\) of zinc with excess silver nitrite to produce silver metal and zinc nitrite. The reaction is stopped before all the zinc metal has reacted and \(29.0 \mathrm{~g}\) of solid metal is present. Calculate the mass of each metal in the 29.0-g mixture.

A \(1.42-\mathrm{g}\) sample of a pure compound, with formula \(\mathrm{M}_{2} \mathrm{SO}_{4}\), was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected, dried, and found to weigh \(1.36 \mathrm{~g}\). Determine the atomic mass of \(\mathrm{M}\), and identify \(\mathrm{M}\).

Chlorisondamine chloride \(\left(\mathrm{C}_{4} \mathrm{H}_{20} \mathrm{Cl}_{6} \mathrm{~N}_{2}\right)\) is a drug used in the treatment of hypertension. A \(1.28-\mathrm{g}\) sample of a medication containing the drug was treated to destroy the organic material and to release all the chlorine as chloride ion. When the filtered solution containing chloride ion was treated with an excess of silver nitrate, \(0.104 \mathrm{~g}\) silver chloride was recovered. Calculate the mass percent of chlorisondamine chloride in the medication, assuming the drug is the only source of chloride.

Describe how you would prepare \(2.00 \mathrm{~L}\) of each of the following solutions. a. \(0.250 \mathrm{M} \mathrm{NaOH}\) from solid \(\mathrm{NaOH}\) b. \(0.250 M \mathrm{NaOH}\) from \(1.00 M \mathrm{NaOH}\) stock solution c. \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) from solid \(\mathrm{K}_{2} \mathrm{CrO}_{4}\) d. \(0.100 M \mathrm{~K}_{2} \mathrm{CrO}_{4}\) from \(1.75 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) stock solution

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.