Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 53

Describe how you would prepare each of the following aqueous solutions, starting with solid KBr: (a) 0.75 L of \(1.5 \times 10^{-2} M \mathrm{KBr},\) (b) 125 \(\mathrm{g}\) of \(0.180 \mathrm{m} \mathrm{KBr},(\mathbf{c}) 1.85 \mathrm{L}\) of a solution that is 12.0\(\% \mathrm{KBr}\) by mass (the density of the solution is 1.10 \(\mathrm{g} / \mathrm{mL}\) , ( \(\mathrm{d}\) ) a 0.150 \(\mathrm{M}\) solution of \(\mathrm{KBr}\) that contains just enough KBr to precipitate 16.0 \(\mathrm{g}\) of AgBr from a solution containing 0.480 mol of \(\mathrm{AgNO}_{3} .\)

Short Answer

Expert verified
(a) Dissolve 1.33625 g of solid KBr in enough water to make a 0.75 L solution of \(1.5 \times 10^{-2} M\) KBr. (b) Dissolve 2.68125 g of solid KBr in 122.31875 g of water to make a 0.180 m KBr aqueous solution of 125 g. (c) Dissolve 244.2 g of solid KBr in 1790.8 g of water to make 1.85 L of a 12.0% by mass KBr solution. (d) Dissolve 0.10161 mol of KBr in enough water to make a 0.6774 L of a 0.150 M KBr solution. This solution contains enough KBr to precipitate 16.0 g of AgBr from a solution containing 0.480 mol of AgNO3.

Step by step solution

01

(a) 1.5x10-2 M KBr Solution

First, find the moles of KBr needed using the concentration and volume of the solution: Moles of KBr = Molarity * Volume = \(1.5 \times 10^{-2} M \times 0.75 L \) = 0.01125 mol Next, convert the moles to mass using the molar mass of KBr (39.10 g/mol for K and 79.90 g/mol for Br): Mass of KBr = Moles of KBr * Molar Mass = 0.01125 mol * (39.10 g/mol + 79.90 g/mol) = 1.33625 g Dissolve 1.33625 g of solid KBr in enough water to make a 0.75 L solution.
02

(b) 0.180 m KBr Solution

First, find the moles of KBr needed using the molality and mass of the solution: Moles of KBr = molality * mass of the solution = \(0.180 \frac{mol}{kg} \times 0.125 kg\) = 0.0225 mol Next, convert the moles to mass using the molar mass of KBr: Mass of KBr = Moles of KBr * Molar Mass = 0.0225 mol * (39.10 g/mol + 79.90 g/mol) = 2.68125 g Now, calculate the mass of water required: Mass of water = Mass of solution - Mass of KBr = 125 g - 2.68125 g = 122.31875 g Dissolve 2.68125 g of solid KBr in 122.31875 g of water to make a 0.180 m KBr aqueous solution of 125 g.
03

(c) 12.0% KBr Solution

First, find the total mass of the solution using the volume and density: Mass of solution = Volume * Density = 1.85 L * 1.10 g/mL * \( \frac{1000 mL}{1 L} \) = 2035 g Next, calculate the mass of KBr and water required using mass percentages: Mass of KBr = mass percent * mass of the solution = 0.12 * 2035 g = 244.2 g Mass of water = Mass of solution - Mass of KBr = 2035 g - 244.2 g = 1790.8 g Dissolve 244.2 g of solid KBr in 1790.8 g of water to make 1.85 L of a 12.0% by mass KBr solution.
04

(d) 0.150 M KBr Solution

First, use stoichiometry to calculate the moles of KBr required to precipitate 16.0 g of AgBr: 1 mol AgBr = 1 mol KBr (1:1 stoichiometry) Moles of KBr = moles of AgBr = \( \frac{16.0 g}{107.87 g/mol + 79.90 g/mol} \) = 0.10161 mol Next, find the volume of the solution needed using the molarity: Volume of solution = moles of KBr / M = \( \frac{0.10161 mol}{0.150 M} \) = 0.6774 L Dissolve the calculated moles (0.10161 mol) of KBr in enough water to make a 0.6774 L of a 0.150 M KBr solution. This solution contains enough KBr to precipitate 16.0 g of AgBr from a solution containing 0.480 mol of AgNO3.

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 ionic compound has a very negative \(\Delta H_{\text { soln in water. }}\) (a) Would you expect it to be very soluble or nearly insoluble in water? (b) Which term would you expect to be the largest negative number: \(\Delta H_{\text { solvent }} \Delta H_{\text { solute }}\) or \(\Delta H_{\text { mix }}\) ?

During a person's typical breathing cycle, the \(\mathrm{CO}_{2}\) concentration in the expired air rises to a peak of 4.6\(\%\) by volume. (a) Calculate the partial pressure of the CO \(_{2}\) in the expired air at its peak, assuming 1 atm pressure and a body temperature of \(37^{\circ} \mathrm{C}\) (b) What is the molarity of the \(\mathrm{CO}_{2}\) in the expired air at its peak, assuming a body temperature of \(37^{\circ} \mathrm{C} ?\)

The Henry's law constant for helium gas in water at \(30^{\circ} \mathrm{C}\) is \(3.7 \times 10^{-4} \mathrm{M} / \mathrm{atm}\) and the constant for \(\mathrm{N}_{2}\) at \(30^{\circ} \mathrm{C}\) is \(6.0 \times 10^{-4} \mathrm{M} / \mathrm{atm}\) . If the two gases are each present at 1.5 atm pressure, calculate the solubility of each gas.

Proteins can be precipitated out of aqueous solution by the addition of an electrolyte; this process is called "salting out" the protein. (a) Do you think that all proteins would be precipitated out to the same extent by the same concentration of the same electrolyte? (b) If a protein has been salted out, are the protein-protein interactions stronger or weaker than they were before the electrolyte was added? (c) A friend of yours who is taking a biochemistry class says that salting out works because the waters of hydration that surround the protein prefer to surround the electrolyte as the electrolyte is added; therefore, the protein's hydration shell is stripped away, leading to protein precipitation. Another friend of yours in the same biochemistry class says that salting out works because the incoming ions adsorb tightly to the protein, making ion pairs on the protein surface, which end up giving the protein a zero net charge in water and therefore leading to precipitation. Discuss these two hypotheses. What kind of measurements would you need to make to distinguish between these two hypotheses?

A "canned heat" product used to warm buffet dishes consists of a homogeneous mixture of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and paraffin, which has an average formula of \(\mathrm{C}_{24} \mathrm{H}_{50}\). What mass of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) should be added to \(620 \mathrm{~kg}\) of the paraffin to produce 8 torr of ethanol vapor pressure at \(35^{\circ} \mathrm{C}\) ? The vapor pressure of pure ethanol at \(35^{\circ} \mathrm{C}\) is 100 torr.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

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