Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A mouthwash contains \(22.5 \%(\mathrm{v} / \mathrm{v})\) alcohol. If the bottle of mouthwash contains \(355 \mathrm{~mL}\), what is the volume, in milliliters, of alcohol?

Short Answer

Expert verified
79.875 mL

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Understand the problem

The problem is asking for the volume of alcohol in a 355 mL bottle of mouthwash. The mouthwash contains 22.5% alcohol by volume (v/v), meaning 22.5% of the total volume is alcohol.
02

Convert the percentage to decimal

Convert the 22.5% to a decimal by dividing by 100. \[ 22.5\text{\textdiv}100 = 0.225 \ \therefore\text{Alcohol percentage as a decimal} = 0.225 \ \text{ } \]
03

Calculate the volume of alcohol

To find the volume of alcohol, multiply the total volume of the mouthwash by the decimal form of the percentage. \[ V_{\text{alcohol}} = \text{total volume} \times \text{alcohol percentage} \ V_{\text{alcohol}} = 355\text{ mL} \times 0.225 \ \therefore V_{\text{alcohol}} = 79.875\text{ mL} \]

Key Concepts

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

Percentage Concentration
Understanding percentage concentration is essential in solving chemistry problems related to solutions. The percentage concentration, symbolized as % (v/v), denotes the volume of a solute (in this case, alcohol) in a given volume of solution (mouthwash). When a solution is marked with a percentage concentration (v/v), it represents how much volume of the solute is present in 100 milliliters of the solution.

For example, a 22.5% (v/v) alcohol solution means there are 22.5 mL of alcohol in every 100 mL of the solution. To make calculations easier, this percentage is often converted into a decimal by dividing by 100. In this case, 22.5% becomes 0.225.

This conversion helps in performing further calculations to determine the precise volume of solute in varying volumes of solutions. Percentage concentration is widely used in preparing solutions in labs, healthcare, and various industrial processes.
Volume Calculation
The next key concept is volume calculation. When dealing with solutions, you often need to determine the volume of one of the components given certain data.

For instance, given the volume and percentage concentration of a solution, we can calculate the volume of the solute. The essential formula to remember is:
\[ V_{\text{solute}} = V_{\text{solution}} \times \text{concentration} \] In this formula, \[ V_{\text{solute}} \] refers to the volume of the solute (here, alcohol), \[ V_{\text{solution}} \] is the volume of the entire solution (mouthwash), and concentration is the decimal form of the percentage.

By understanding and applying this formula, you can accurately calculate the amount of any component within a mixed solution. This process is particularly useful when it comes to preparing exact formulations in pharmaceuticals, beverages, and other industries.
Alcohol Solution
Alcohol solutions are a common topic in chemistry. These solutions can be applied in various contexts, such as medical products, beverages, and sanitizing agents.

In our given exercise, the mouthwash solution contains alcohol as a solute. The concentration of alcohol is critical both for effectiveness and safety. The volume of alcohol can be calculated by applying the percentage concentration of the solution.

For mouthwashes with 22.5% alcohol concentration, this means the solution consists of 22.5 parts alcohol and 77.5 parts other components by volume per 100 parts total. To find the exact volume of alcohol in a 355 mL bottle, you multiply 355 by the converted decimal concentration (0.225). This gives a final volume of 79.875 mL of alcohol in the mouthwash solution.

Understanding how to calculate and work with alcohol solutions is important in fields ranging from healthcare to food science. By mastering these calculations, you can ensure proper formulation and utilization of solutions in various applications.

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

Calculate the amount of solution (g or mL) that contains each of the following amounts of solute: a. \(5.0 \mathrm{~g}\) of \(\mathrm{LiNO}_{3}\) from a \(25 \%(\mathrm{~m} / \mathrm{m}) \mathrm{Li} \mathrm{NO}_{3}\) solution b. \(40.0 \mathrm{~g}\) of \(\mathrm{KOH}\) from a \(10.0 \%(\mathrm{~m} / \mathrm{v}) \mathrm{KOH}\) solution c. \(2.0 \mathrm{~mL}\) of formic acid from a \(10.0 \%(\mathrm{v} / \mathrm{v})\) formic acid solution

Calculate the mass percent \((\mathrm{m} / \mathrm{m})\) for the solute in each of the following solutions: a. \(25 \mathrm{~g}\) of \(\mathrm{KCl}\) and \(125 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) b. \(12 \mathrm{~g}\) of sugar in \(225 \mathrm{~g}\) of tea solution with sugar c. \(8.0 \mathrm{~g}\) of \(\mathrm{CaCl}_{2}\) in \(80.0 \mathrm{~g}\) of \(\mathrm{CaCl}_{2}\) solution

A solution contains \(40 . \mathrm{mEq} / \mathrm{L}\) of \(\mathrm{Cl}^{-}\) and \(15 \mathrm{mEq} / \mathrm{L}\) of \(\mathrm{HPO}_{4}{ }^{2-}\). If \(\mathrm{Na}^{+}\) is the only cation in the solution, what is the Na concentration, in milliequivalents per liter?

Calculate the molarity of the solution when water is added to prepare each of the following: a. \(25.0 \mathrm{~mL}\) of a \(18.0 \mathrm{M} \mathrm{HCl}\) solution is diluted to \(500 . \mathrm{mL}\) b. \(50.0 \mathrm{~mL}\) of a \(1.50 \mathrm{M} \mathrm{NaCl}\) solution is diluted to \(125 \mathrm{~mL}\) c. \(4.50 \mathrm{~mL}\) of a \(8.50 \mathrm{M} \mathrm{KOH}\) solution is diluted to \(75.0 \mathrm{~mL}\)

Write a balanced equation for the dissociation of each of the following strong electrolytes in water: a. \(\mathrm{KCl}\) b. \(\mathrm{CaCl}_{2}\) c. \(\mathrm{K}_{3} \mathrm{PO}_{4}\) d. \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\)

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.

Sign-up for free