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

What is the difference between a \(5 \%(\mathrm{~m} / \mathrm{m})\) glucose solution and a \(5 \%(\mathrm{~m} / \mathrm{v})\) glucose solution?

Short Answer

Expert verified
5% (m/m) is by mass; 5% (m/v) is by volume.

Step by step solution

01

Understand \text{%(m/m)}

A 5% (m/m) glucose solution means that there are 5 grams of glucose in every 100 grams of the solution. This is a mass/mass percentage.
02

Understand \text{%(m/v)}

A 5% (m/v) glucose solution means that there are 5 grams of glucose in every 100 milliliters of the solution. This is a mass/volume percentage.
03

Compare the Two Solutions

The key difference is that %(m/m) is based on the mass of the entire solution, whereas %(m/v) is based on the volume of the solution. This means that for a 5% (m/m) solution, the concentration is measured by mass, and for a 5% (m/v) solution, the concentration is measured by volume.
04

Implications of the Difference

The mass/mass solution means if you have 100 grams of solution, it includes the mass of glucose within it. For a mass/volume solution, the solution might have a different density and the total mass will vary depending on the volume (which could be 100 mL but the mass isn’t specified).

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 Vaia!

Key Concepts

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

mass/mass percentage
Mass/mass percentage, often written as \(m/m\), is a way to express the concentration of a solution. It tells you how many grams of a solute are present in 100 grams of the solution. For example, a 5% \(m/m\) glucose solution means there are 5 grams of glucose in every 100 grams of the solution.
  • The solute is the substance that is being dissolved, in this case, glucose.
  • The solvent is the substance doing the dissolving, typically water in many biological solutions.
This percentage makes it easy to visualize how much solute is present without needing additional information about the solution's volume. Thus, this measurement is particularly useful when the weights of solute and solvent are easily obtainable.
mass/volume percentage
Mass/volume percentage, often denoted as \(m/v\), describes the concentration of a solution as the mass of the solute present in a certain volume of the solution. For instance, a 5% \(m/v\) glucose solution indicates there are 5 grams of glucose per 100 milliliters of solution.
Unlike the mass/mass percentage, this measurement requires knowing the solution's volume.
  • The solute is measured by its mass (grams) while the solvent is indicated by volume (milliliters).
  • This measurement is useful in practical laboratory settings where volumes are easier to measure accurately than weights.
Overall, this percentage is highly beneficial for solutions where the volume is a critical component, such as in medical or pharmacological applications.
solution density
Solution density refers to the mass of the solution per unit of volume, usually expressed in grams per milliliter (g/mL). This value varies depending on the solution's composition.
  • If you have a high density, it means the solution has more mass packed into each milliliter.
  • A low-density solution has less mass per milliliter.
Knowing the density can help you convert between \(m/m\) and \(m/v\) percentages. For example, if you know the density of a 5% \(m/m\) glucose solution, you can calculate how many grams per milliliter are present and vice versa. Density is a bridge between mass and volume calculations, reflecting how compact the solute particles are within the solvent.
concentration measurements
Concentration measurements are crucial in chemistry and many scientific disciplines. These measurements tell you how much solute is in a certain amount of solvent or solution.
There are various ways to express concentration:
  • The mass/mass percentage (\((m/m)\)) and mass/volume percentage (\((m/v)\)) are two common methods.
  • Other methods include molarity, molality, and normality, which involve different unit measurements and contexts.
Accurate concentration measurements are essential for experiments, chemical reactions, and even medical treatments, where precise amounts of substances must be used. They help ensure that reactions occur correctly, predict outcomes, and maintain safety standards.

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 concentration of each of the following diluted solutions: a. \(1.0 \mathrm{~L}\) of a \(4.0 \mathrm{M} \mathrm{HNO}_{3}\) solution is added to water so that the final volume is \(8.0 \mathrm{~L}\). b. Water is added to \(0.25 \mathrm{~L}\) of a \(6.0 \mathrm{M}\) NaF solution to make 2.0 L of a diluted NaF solution. c. A \(50.0-\mathrm{mL}\) sample of an \(8.0 \%(\mathrm{~m} / \mathrm{v}) \mathrm{KBr}\) solution is diluted with water so that the final volume is \(200.0 \mathrm{~mL}\). d. A \(5.0\) -mL sample of a \(50.0 \%\) (m/v) acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) solution is added to water to give a final volume of \(25 \mathrm{~mL}\).

A solution is prepared by dissolving \(22.0 \mathrm{~g}\) of \(\mathrm{NaOH}\) in \(118.0 \mathrm{~g}\) of water. The \(\mathrm{NaOH}\) solution has a density of \(1.15 \mathrm{~g} / \mathrm{mL}\). a. What is the mass percent \((\mathrm{m} / \mathrm{m})\) of the \(\mathrm{NaOH}\) solution? b. What is the total volume (mL) of the solution? c. What is the molarity (M) of the solution?

How many grams of \(\mathrm{K}_{2} \mathrm{CO}_{3}\) are in \(750 \mathrm{~mL}\) of a \(3.5 \%(\mathrm{~m} / \mathrm{v})\) \(\mathrm{K}_{2} \mathrm{CO}_{3}\) solution?

How many milliliters of each of the following solutions will provide \(25.0 \mathrm{~g}\) of \(\mathrm{KOH}\) ? a. \(2.50 \mathrm{M} \mathrm{KOH}\) solution b. \(0.750 \mathrm{M}\) KOH solution c. \(5.60 \mathrm{M} \mathrm{KOH}\) solution

Calculate the grams or milliliters of solute needed to prepare the following solutions: a. \(50.0 \mathrm{~mL}\) of a \(5.0 \%(\mathrm{~m} / \mathrm{v}) \mathrm{KCl}\) solution b. \(1250 \mathrm{~mL}\) of a \(4.0 \%(\mathrm{~m} / \mathrm{v}) \mathrm{NH}_{4} \mathrm{Cl}\) solution c. \(250 . \mathrm{mL}\) of a \(10.0 \%(\mathrm{v} / \mathrm{v})\) acetic acid 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.

Sign-up for free