Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 39

Which of the following is a \(0.500 \mathrm{M} \mathrm{KCl}\) solution? (a) \(0.500 \mathrm{g} \mathrm{KCl} / \mathrm{mL}\) solution; (b) \(36.0 \mathrm{g} \mathrm{KCl} / \mathrm{L}\) solu- tion; (c) 7.46 mg KCl/mL solution; (d) 373 g KCl in 10.00 L solution

Short Answer

Expert verified
The 0.500 M KCl solution is option (d) - 373 g KCl in 10.00 L solution.

Step by step solution

01

Calculation for option (a)

For option (a), we are given the mass of KCl in grams per mL of solution. However, neither of these can be directly converted to moles per liter. The data makes no sense and is incorrect for a molar solution.
02

Calculation for option (b)

For option (b), we are given the mass of KCl in grams per litre. To convert this to moles per liter (i.e., molarity), we divide the mass by the molar mass of KCl (which is approximately 74.55 g/mol). That gives \((36.0 \mathrm{g} \mathrm{KCl} / \mathrm{L})/74.55 \mathrm{g/mol} = 0.483 \mathrm{M}\). Therefore, option (b) is not a 0.500 M KCl solution.
03

Calculation for option (c)

For option (c), we are given the mass of KCl in milligrams per mL of solution. We can convert milligrams to grams and mL to L to get the molarity: \((7.46 \mathrm{mg} \mathrm{KCl}/\mathrm{mL}) * \mathrm{1g/1000mg} * \mathrm{1000mL/1L} = 7.46 \mathrm{g} \mathrm{KCl} / \mathrm{L} \). Then, divide this by the molar mass of KCl (74.55 g/mol) to get the molarity: \(7.46 \mathrm{g} / 74.55 \mathrm{g/mol} = 0.100 \mathrm{M}\). Therefore, option (c) is not a 0.500 M KCl solution.
04

Calculation for option (d)

Finally, for option (d), we have 373 g of KCl in 10.00 L of solution. To get molarity, again divide the mass by the molar mass of KCl, and then by the volume of the solution in liters: \( (373 \mathrm{g} / 74.55 \mathrm{g/mol}) / 10.00 \mathrm{L} = 0.500 \mathrm{M}\). Therefore, option (d) is a 0.500 M KCl solution.

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

Appendix E describes a useful study aid known as concept mapping. Using the method presented in Appendix \(\mathrm{E},\) construct a concept map relating the topics found in Sections \(4-3,4-4,\) and \(4-5\).

Write a chemical equation to represent the complete combustion of malonic acid, a compound with \(34.62 \% \mathrm{C}, 3.88 \% \mathrm{H},\) and \(61.50 \% \mathrm{O},\) by mass.

The following set of reactions is to be used as the basis of a method for producing nitric acid, \(\mathrm{HNO}_{3}\) Calculate the minimum masses of \(\mathrm{N}_{2}, \mathrm{H}_{2^{\prime}}\) and \(\mathrm{O}_{2}\) required per kilogram of \(\mathrm{HNO}_{3}\) $$\begin{array}{l} \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g}) \\ 4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) \\ 3 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{NO}(\mathrm{g}) \end{array}$$

What is the molarity of \(\mathrm{NaCl}(\mathrm{aq})\) if a solution has 1.52 ppm Na? Assume that NaCl is the only source of Na and that the solution density is \(1.00 \mathrm{g} / \mathrm{mL}\) (The unit \(p p m\) is parts per million; here it can be taken to mean g Na per million grams of solution.)

When water and methanol, \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l}),\) are mixed, the total volume of the resulting solution is not equal to the sum of the pure liquid volumes. (Refer to Exercise 99 for an explanation.) When \(72.061 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) and \(192.25 \mathrm{g}\) \(\mathrm{CH}_{3} \mathrm{OH}\) are mixed at \(25^{\circ} \mathrm{C},\) the resulting solution has a density of \(0.86070 \mathrm{g} / \mathrm{mL} .\) At \(25^{\circ} \mathrm{C},\) the densities of water and methanol are \(0.99705 \mathrm{g} / \mathrm{mL}\) and \(0.78706\) \(\mathrm{g} / \mathrm{mL},\) respectively. (a) Calculate the volumes of the pure liquid samples and the solution, and show that the pure liquid volumes are not additive. [ Hint: Although the volumes are not additive, the masses are.] (b) Calculate the molarity of \(\mathrm{CH}_{3} \mathrm{OH}\) in this solution.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

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