Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 31

Which of the following solutions of strong electrolytes contains the largest number of moles of chloride ions: \(100.0 \mathrm{~mL}\) of \(0.30 \mathrm{M} \mathrm{AlCl}_{3}, 50.0 \mathrm{~mL}\) of \(0.60 \mathrm{M} \mathrm{MgCl}_{2}\), or \(200.0 \mathrm{~mL}\) of \(0.40 \mathrm{M} \mathrm{NaCl} ?\)

Short Answer

Expert verified
The solution with the largest number of moles of chloride ions is \(100.0 \mathrm{~mL}\) of $0.30 \mathrm{M} \mathrm{AlCl}_{3}$, which contains 0.09 moles of Cl⁻.

Step by step solution

01

Identify the dissociation of the strong electrolytes

Each of the strong electrolytes given will dissociate completely in the solution, producing respective ions. We can write the dissociation reaction as: AlCl₃ ↔ Al³⁺ + 3Cl⁻ MgCl₂ ↔ Mg²⁺ + 2Cl⁻ NaCl ↔ Na⁺ + Cl⁻
02

Calculate the moles of chloride ions produced by each electrolyte

To find the moles of chloride ions produced by each electrolyte, we'll follow this formula: moles of Cl⁻ = [concentration (M) * volume (L)] * (moles of Cl⁻ / moles of electrolyte) Make sure to convert the given volume from mL to L by dividing the volume by 1000.
03

Calculate the moles of chloride ions in the AlCl₃ solution

Using the given concentration and volume of AlCl₃ solution: moles of Cl⁻ = [0.30 M * (100.0 mL / 1000)] * 3 moles of Cl⁻ = 0.09 moles
04

Calculate the moles of chloride ions in the MgCl₂ solution

Using the given concentration and volume of MgCl₂ solution: moles of Cl⁻ = [0.60 M * (50.0 mL / 1000)] * 2 moles of Cl⁻ = 0.06 moles
05

Calculate the moles of chloride ions in the NaCl solution

Using the given concentration and volume of NaCl solution: moles of Cl⁻ = [0.40 M * (200.0 mL / 1000)] * 1 moles of Cl⁻ = 0.08 moles
06

Compare the moles of chloride ions for each solution

Now that we've calculated the moles of chloride ions for each electrolyte solution, we can compare them to see which one has the highest value. AlCl₃: 0.09 moles Cl⁻ MgCl₂: 0.06 moles Cl⁻ NaCl: 0.08 moles Cl⁻ Out of these values, the AlCl₃ solution contains the largest number of moles of chloride ions (0.09 moles). Therefore, \(100.0 \mathrm{~mL}\) of $0.30 \mathrm{M} \mathrm{AlCl}_{3}$ contains the largest number of moles of chloride ions.

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

An average human being has about \(5.0 \mathrm{~L}\) of blood in his or her body. If an average person were to eat \(32.0 \mathrm{~g}\) of sugar (sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}, 342.30 \mathrm{~g} / \mathrm{mol}\) ), and all that sugar were dissolved into the bloodstream, how would the molarity of the blood sugar change?

Zinc and magnesium metal each react with hydrochloric acid according to the following equations: $$ \begin{aligned} \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) & \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) \\ \mathrm{Mg}(s)+2 \mathrm{HCl}(a q) & \longrightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2}(g) \end{aligned} $$ A \(10.00-\mathrm{g}\) mixture of zinc and magnesium is reacted with the stoichiometric amount of hydrochloric acid. The reaction mixture is then reacted with \(156 \mathrm{~mL}\) of \(3.00 \mathrm{M}\) silver nitrate to produce the maximum possible amount of silver chloride. a. Determine the percent magnesium by mass in the original mixture. b. If \(78.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) was added, what was the concentration of the \(\mathrm{HCl} ?\)

You wish to prepare \(1 \mathrm{~L}\) of a \(0.02-M\) potassium iodate solution. You require that the final concentration be within \(1 \%\) of \(0.02 M\) and that the concentration must be known accurately to the fourth decimal place. How would you prepare this solution? Specify the glassware you would use, the accuracy needed for the balance, and the ranges of acceptable masses of \(\mathrm{KIO}_{7}\) that can be used.

Calculate the sodium ion concentration when \(70.0 \mathrm{~mL}\) of 3.0 \(M\) sodium carbonate is added to \(30.0 \mathrm{~mL}\) of \(1.0 \mathrm{M}\) sodium bicarbonate.

A mixture contains only sodium chloride and potassium chloride. A \(0.1586-\mathrm{g}\) sample of the mixture was dissolved in water. It took \(22.90 \mathrm{~mL}\) of \(0.1000 \mathrm{M} \mathrm{AgNO}_{3}\) to completely precipitate all the chloride present. What is the composition (by mass percent) of the mixture?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

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