/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 Pure acetic acid, known as glaci... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 77

Pure acetic acid, known as glacial acetic acid, is a liquid with a density of \(1.049 \mathrm{~g} / \mathrm{mL}\) at \(25^{\circ} \mathrm{C}\). Calculate the molarity of a solution of acetic acid made by dissolving \(20.00 \mathrm{~mL}\) of glacial acetic acid at \(25^{\circ} \mathrm{C}\) in enough water to make \(250.0 \mathrm{~mL}\) of solution.

Short Answer

Expert verified
The molarity of the acetic acid solution is 1.396 M.

Step by step solution

01

Calculate the mass of acetic acid using its density and volume

To calculate the mass of acetic acid in the given volume, we use the formula: mass = density 脳 volume The volume of glacial acetic acid is 20.00 mL, and its density is 1.049 g/mL. Plugging in the values, we get: mass = 1.049 g/mL 脳 20.00 mL mass = 20.98 g So, the mass of acetic acid dissolved is 20.98 g.
02

Convert the mass of acetic acid to moles using its molar mass

To find the moles of acetic acid, we will use the molar mass of acetic acid: molar mass of acetic acid = 60.05 g/mol Now, we can find the moles by dividing the mass of acetic acid by its molar mass: moles of acetic acid = mass / molar mass moles of acetic acid = 20.98 g / 60.05 g/mol moles of acetic acid = 0.349 mol We have 0.349 moles of acetic acid.
03

Calculate the molarity using the moles of acetic acid and the final volume of the solution

Finally, we can calculate the molarity of the solution using the moles of acetic acid and the final volume of the solution: Molarity (M) = moles of solute / volume of solution (in liters) The final volume of the solution is 250.0 mL, which we need to convert to liters: 250.0 mL = 250.0 / 1000 L = 0.250 L Now, we can calculate the molarity of the solution: Molarity = 0.349 mol / 0.250 L Molarity = 1.396 mol/L Therefore, the molarity of the acetic acid solution is 1.396 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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

Using modern analytical techniques, it is possible to detect sodium ions in concentrations as low as \(50 \mathrm{pg} / \mathrm{mL}\). What is this detection limit expressed in (a) molarity of \(\mathrm{Na}^{+}\) (b) \(\mathrm{Na}^{+}\) ions per cubic centimeter?

Determine the oxidation number of sulfur in each of the following substances: (a) barium sulfate, \(\mathrm{BaSO}_{4},\) (b) sulfurous acid, \(\mathrm{H}_{2} \mathrm{SO}_{3},\) (c) strontium sulfide, \(\mathrm{SrS},\) (d) hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S} .\) (e) Based on these compounds what is the range of oxidation numbers seen for sulfur? Is there any relationship between the range of accessible oxidation states and sulfur's position on the periodic table?

The mass percentage of chloride ion in a \(25.00-\mathrm{mL}\) sample of seawater was determined by titrating the sample with silver nitrate, precipitating silver chloride. It took \(42.58 \mathrm{~mL}\) of \(0.2997 \mathrm{M}\) silver nitrate solution to reach the equivalence point in the titration. What is the mass percentage of chloride ion in the seawater if its density is \(1.025 \mathrm{~g} / \mathrm{mL} ?\)

A solution of \(100.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{KOH}\) is mixed with a solution of \(200.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{NiSO}_{4}\). (a) Write the balanced chemical equation for the reaction that occurs. (b) What precipitate forms? (c) What is the limiting reactant? (d) How many grams of this precipitate form? (e) What is the concentration of each ion that remains in solution?

Indicate the concentration of each ion present in the solution formed by mixing (a) \(42.0 \mathrm{~mL}\) of \(0.170 \mathrm{M} \mathrm{NaOH}\) and \(37.6 \mathrm{~mL}\) of \(0.400 \mathrm{M} \mathrm{NaOH}\), (b) \(44.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(25.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{KCl},(\mathrm {c}) 3.60 \mathrm{~g} \mathrm{KCl}\) in \(75.0 \mathrm{~mL}\) of \(0.250 \mathrm{M}\) \(\mathrm{CaCl}_{2}\) solution. Assume that the volumes are additive.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

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.