Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 98

A \(60.0-\mathrm{mL} 0.513 \mathrm{M}\) glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) solution is mixed with \(120.0 \mathrm{~mL}\) of \(2.33 \mathrm{M}\) glucose solution. What is the concentration of the final solution? Assume the volumes are additive.

Short Answer

Expert verified
The concentration of the glucose solution after mixing is 1.724 M.

Step by step solution

01

Calculate Moles in Each Solution

The concentration of a solution tells us how many moles of solute are present in every liter of solution. By multiplying the volume of solution (converting it from milliliters to liters) by the concentration, the number of moles of glucose in each solution can be calculated. For the 60.0 mL of 0.513 M glucose solution: moles = 0.060 L * 0.513 mol/L = 0.03078 mol and For the 120.0 mL of 2.33 M glucose solution: moles = 0.120 L * 2.33 mol/L = 0.2796 mol
02

Calculate Total Moles and Total Volume

The total number of moles of glucose is the sum of the moles in each solution (0.03078 mol + 0.2796 mol = 0.31038 mol). The total volume of the mixture is also the sum of the volumes of each solution (60 ml + 120 ml = 180 ml = 0.18 L).
03

Calculate Final Concentration

The concentration of the final solution can now be calculated by dividing the total number of moles of glucose by the total volume of the mixture in liters: concentration = moles/volume = 0.31038 mol / 0.18 L = 1.724 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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

List the general properties of acids and bases.

Classify these reactions according to the types discussed in the chapter: (a) \(\mathrm{Cl}_{2}+2 \mathrm{OH}^{-} \longrightarrow \mathrm{Cl}^{-}+\mathrm{ClO}^{-}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Ca}^{2+}+\mathrm{CO}_{3}^{2-} \longrightarrow \mathrm{CaCO}_{3}\) (c) \(\mathrm{NH}_{3}+\mathrm{H}^{+} \longrightarrow \mathrm{NH}_{4}^{+}\) (d) \(2 \mathrm{CCl}_{4}+\mathrm{CrO}_{4}^{2-} \longrightarrow\) \(2 \mathrm{COCl}_{2}+\mathrm{CrO}_{2} \mathrm{Cl}_{2}+2 \mathrm{Cl}^{-}\) (e) \(\mathrm{Ca}+\mathrm{F}_{2} \longrightarrow \mathrm{CaF}_{2}\) (f) \(2 \mathrm{Li}+\mathrm{H}_{2} \longrightarrow 2 \mathrm{LiH}\) (g) \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{Na}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{NaNO}_{3}+\mathrm{BaSO}_{4}\) (h) \(\mathrm{CuO}+\mathrm{H}_{2} \longrightarrow \mathrm{Cu}+\mathrm{H}_{2} \mathrm{O}\) (i) \(\mathrm{Zn}+2 \mathrm{HCl} \longrightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2}\) (j) \(2 \mathrm{FeCl}_{2}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{FeCl}_{3}\)

(a) Without referring to Figure 4.10 , give the oxidation numbers of the alkali and alkaline earth metals in their compounds. (b) Give the highest oxidation numbers that the Groups \(3 \mathrm{~A}-7 \mathrm{~A}\) elements can have.

A 325-mL sample of solution contains \(25.3 \mathrm{~g}\) of \(\mathrm{CaCl}_{2}\). (a) Calculate the molar concentration of \(\mathrm{Cl}^{-}\) in this solution. (b) How many grams of \(\mathrm{Cl}^{-}\) are in \(0.100 \mathrm{~L}\) of this solution?

Write the equation that enables us to calculate the concentration of a diluted solution. Give units for all the terms.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation
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