Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 124

Calculate the concentration of all ions present when 0.160 \(\mathrm{g}\) of \(\mathrm{MgCl}_{2}\) is dissolved in 100.0 \(\mathrm{mL}\) of solution.

Short Answer

Expert verified
When 0.160 g of MgCl₂ is dissolved in 100.0 mL of solution, the concentration of Mg²⁺ ions is 0.0168 M, and the concentration of Cl⁻ ions is 0.0336 M.

Step by step solution

01

Convert grams of MgCl₂ to moles.

To find the moles of MgCl₂ dissolved, we must first convert the 0.160 g of MgCl₂ to moles. Using the molar mass of MgCl₂, which can be found by adding the molar masses of Mg (24.31 g/mol) and two Cl atoms (2 x 35.45 g/mol): Molar mass of MgCl₂ = 24.31 g/mol + (2 x 35.45 g/mol) = 95.21 g/mol Now, convert the mass of MgCl₂ to moles using the molar mass: Moles of MgCl₂ = (0.160 g) / (95.21 g/mol) = 0.00168 mol
02

Calculate the moles of Mg²⁺ and Cl⁻ ions produced.

When MgCl₂ dissolves, it produces one mole of Mg²⁺ ions and two moles of Cl⁻ ions for each mole of MgCl₂. Therefore, we can find the moles of Mg²⁺ and Cl⁻ ions produced as follows: Moles of Mg²⁺ = Moles of MgCl₂ = 0.00168 mol Moles of Cl⁻ = 2 x Moles of MgCl₂ = 2 x 0.00168 mol = 0.00336 mol
03

Convert the volume of the solution to liters.

The volume of the solution is given in milliliters (mL), and we need to convert it to liters (L) to calculate the molar concentration. To do this, divide the volume in milliliters by 1000: Volume in liters = 100.0 mL / 1000 = 0.100 L
04

Calculate the concentration of Mg²⁺ and Cl⁻ ions.

Now we can find the concentration of Mg²⁺ and Cl⁻ ions in the solution by dividing the moles of each ion by the volume of the solution in liters: Concentration of Mg²⁺ = Moles of Mg²⁺ / Volume in liters = 0.00168 mol / 0.100 L = 0.0168 M Concentration of Cl⁻ = Moles of Cl⁻ / Volume in liters = 0.00336 mol / 0.100 L = 0.0336 M Therefore, when 0.160 g of MgCl₂ is dissolved in 100.0 mL of solution, the concentration of Mg²⁺ ions is 0.0168 M, and the concentration of Cl⁻ ions is 0.0336 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

Specify which of the following are oxidation–reduction reactions, and identify the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced. a. $\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \rightarrow 2 \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)$ b. $\mathrm{HCl}(g)+\mathrm{NH}_{3}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)$ c. $\mathrm{SiCl}_{4}(i)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 4 \mathrm{HCl}(a q)+\mathrm{SiO}_{2}(s)$ d. $\mathrm{SiCl}_{4}(l)+2 \mathrm{Mg}(s) \rightarrow 2 \mathrm{MgCl}_{2}(s)+\mathrm{Si}(s)$ e. $\mathrm{Al}(\mathrm{OH})_{4}-(a q) \rightarrow \mathrm{AlO}_{2}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$

Chlorisondamine chloride $\left(\mathrm{C}_{14} \mathrm{H}_{20} \mathrm{Cl}_{6} \mathrm{N}_{2}\right)$ is a drug used in the treatment of hypertension. A 1.28-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 g silver chloride was recovered. Calculate the mass percent of chlorisondamine chloride in the medication, assuming the drug is the only source of chloride.

A sample may contain any or all of the following ions: \(\mathrm{Hg}_{2}^{2+}\) \(\mathrm{Ba}^{2+}\) and \(\mathrm{Mn}^{2+}\) a. No precipitate formed when an aqueous solution of NaCl was added to the sample solution. b. No precipitate formed when an aqueous solution of $\mathrm{Na}_{2} \mathrm{SO}_{4}$ was added to the sample solution. c. A precipitate formed when the sample solution was made basic with NaOH. Which ion or ions are present in the sample solution?

Consider the reaction of 19.0 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 g of solid metal is present. Calculate the mass of each metal in the 29.0-g mixture

The thallium (present as \(\mathrm{Tl}_{2} \mathrm{SO}_{4} )\) in a \(9.486-\mathrm{g}\) pesticide sam- ple was precipitated as thallium(I) iodide. Calculate the mass percent of \(\mathrm{TI}_{2} \mathrm{SO}_{4}\) in the sample if 0.1824 \(\mathrm{g}\) of TIII was recovered.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Physical 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