Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 49

The density of acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) is 0.786 \(\mathrm{g} / \mathrm{mL}\) and the density of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is 0.791 \(\mathrm{g} / \mathrm{mL} . \mathrm{A}\) solution is made by dissolving 22.5 \(\mathrm{mL}\) of \(\mathrm{CH}_{3} \mathrm{OH}\) in 98.7 \(\mathrm{mL}\) of \(\mathrm{CH}_{3} \mathrm{CN}\) . (a) What is the mole fraction of methanol in the solution? (b) What is the molality of the solution? (c) Assuming that the volumes are additive, what is the molarity of \(\mathrm{CH}_{3} \mathrm{OH}\) in the solution?

Short Answer

Expert verified
To find the mole fraction, molality, and molarity of methanol (CH₃OH) in the given solution, follow these steps: 1. Calculate the mass of each component using their densities and volumes. 2. Calculate the moles of each component using their molar masses. 3. Calculate the mole fraction of methanol by dividing its moles by the total moles of the solution. 4. Calculate the molality of methanol by dividing its moles by the mass of the solvent (acetonitrile, CH₃CN) in kilograms. 5. Calculate the molarity of methanol by dividing its moles by the total volume of the solution in liters.

Step by step solution

01

Calculate the mass of each component

Since we know the density and volume of both acetonitrile (CH₃CN) and methanol (CH₃OH), we can determine the mass of each component by using the formula: mass = density * volume Mass of acetonitrile (CH₃CN): \( mass_{CH_3 CN} = density_{CH_3 CN} * volume_{CH_3 CN} \) Mass of methanol (CH₃OH): \( mass_{CH_3 OH} = density_{CH_3 OH} * volume_{CH_3 OH} \)
02

Calculate the moles of each component

To find the moles, we can use the molar mass of each substance: Moles of acetonitrile (CH₃CN): \( moles_{CH_3 CN} = \frac{mass_{CH_3 CN}}{molar\, mass_{CH_3 CN}} \) Moles of methanol (CH₃OH): \( moles_{CH_3 OH} = \frac{mass_{CH_3 OH}}{molar\, mass_{CH_3 OH}} \)
03

Calculate the mole fraction of methanol (CH₃OH)

The mole fraction of methanol can be found by dividing the moles of methanol by the total moles of the solution: Mole fraction of methanol (CH₃OH): \( X_{CH_3 OH} = \frac{moles_{CH_3 OH}}{moles_{CH_3 OH} + moles_{CH_3 CN}} \)
04

Calculate the molality of the methanol (CH₃OH) solution

Molality is the ratio of moles of solute (CH₃OH) to the mass of the solvent (CH₃CN) in kilograms. Molality of methanol (CH₃OH) in the solution: \( molality_{CH_3 OH} = \frac{moles_{CH_3 OH}}{mass_{CH_3 CN}(kg)} \)
05

Calculate the molarity of methanol (CH₃OH) in the solution

Molarity is the ratio of moles of solute (CH₃OH) to the total volume of the solution in liters. Since the volumes of CH₃OH and CH₃CN are given, we can find the total volume and convert it to liters. Total volume of the solution in liters: \( V_{total} = \frac{volume_{CH_3 OH} + volume_{CH_3 CN}}{1000} \) Molarity of methanol (CH₃OH) in the solution: \( molarity_{CH_3 OH} = \frac{moles_{CH_3 OH}}{V_{total}} \)

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

The solubility of MnSO_ \(\cdot \mathrm{H}_{2} \mathrm{O}\) in water at \(20^{\circ} \mathrm{C}\) is 70 \(\mathrm{g}\) per 100 \(\mathrm{mL}\) of water. (a) Is a 1.22 \(\mathrm{M}\) solution of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) in water at \(20^{\circ} \mathrm{C}\) saturated, supersaturated, or unsaturated? (b) Given a solution of MnSO_ \(\cdot \mathrm{H}_{2} \mathrm{O}\) of unknown concentration, what experiment could you perform to determine whether the new solution is saturated, supersaturated, or unsaturated?

Would you expect alanine (an amino acid) to be more soluble in water or in hexane?

The partial pressure of \(\mathrm{O}_{2}\) in air at sea level is 0.21 atm. Using the data in Table \(13.1,\) together with Henry's law, calculate the molar concentration of \(\mathrm{O}_{2}\) in the surface water of a mountain lake saturated with air at \(20^{\circ} \mathrm{C}\) and an atmospheric pressure of 650 torr.

A sulfuric acid solution containing 571.6 \(\mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) per liter of solution has a density of 1.329 \(\mathrm{g} / \mathrm{cm}^{3} .\) Calculate (a) the mass percentage, (b) the mole fraction, (c) the molality, ( \mathbf{d} ) ~ t h e ~ m o l a r i t y ~ o f ~ \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in this solution.

Soaps consist of compounds such as sodium stearate, \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{16} \mathrm{COO}-\mathrm{Na}^{+},\) that have both hydrophobic and hydrophilic parts. Consider the hydrocarbon part of sodium stearate to be the "tail" and the charged part to be the "head." (a) Which part of sodium stearate, head or tail, is more likely to be solvated by water? (b) Grease is a complex mixture of (mostly) hydrophobic compounds. Which part of sodium stearate, head or tail, is most likely to bind to grease? (c) If you have large deposits of grease that you want to wash away with water, you can see that adding sodium stearate will help you produce an emulsion. What intermolecular interactions are responsible for this?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Organic 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