Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 32

Calculate the mole fraction of solute in the following aqueous solutions: (a) \(21.7 \% \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH},\) by mass; (b) \(0.684 m \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\) (urea).

Short Answer

Expert verified
The calculations following the above steps will give you the mole fractions of ethanol and urea in their respective solutions. Due to varying molar masses of the components, the results will differ. Always use accurate molar masses for precise results.

Step by step solution

01

Calculate moles of solute and solvent for (a)

First, assume you have 100g of the solution. This means there are \(21.7 g\) of ethanol and \(100g - 21.7g = 78.3g\) of water (solvent). Next, convert the mass of each component to moles by dividing by their respective molar mass. The molar mass of ethanol \(CH3CH2OH\) is approximately \(46.07 g/mol\) and for water, it is \(18.015 g/mol\). So, you get \[\text{moles of } CH3CH2OH = \frac{21.7g}{46.07 g/mol} \]and \[\text{moles of } H2O = \frac{78.3g}{18.015 g/mol} \]
02

Calculate mole fraction for (a)

The mole fraction of a component in a solution is given by the number of moles of that component divided by the total number of moles of all components. So, the mole fraction of ethanol is \[\text{mole fraction of } CH3CH2OH = \frac{\text{moles of } CH3CH2OH}{\text{moles of } CH3CH2OH + \text{moles of } H2O} \]
03

Calculate moles of solute and solvent for (b)

In this step, understand that molality is defined as the number of moles of solute per kilogram of solvent. That means, in a 0.684 molal solution, there are 0.684 moles of urea for every kg (or 1000 grams) of water. So, calculate the moles of each component. You have 0.684 moles of \(CO(NH2)2\) and \[\text{moles of } H2O = \frac{1000g}{18.015 g/mol} \]
04

Calculate mole fraction for (b)

The mole fraction of urea in the solution is \[\text{mole fraction of } CO(NH2)2 = \frac{\text{moles of } CO(NH2)2}{\text{moles of } CO(NH2)2 + \text{moles of } H2O} \]

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

What are the partial and total vapor pressures of a solution obtained by mixing 35.8 g benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\) and \(56.7 \mathrm{g}\) toluene, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3},\) at \(25^{\circ} \mathrm{C} ? \mathrm{At} 25^{\circ} \mathrm{C}\) the vapor pressure of \(\mathrm{C}_{6} \mathrm{H}_{6}=95.1 \mathrm{mm} \mathrm{Hg} ;\) the vapor pressure of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}=28.4 \mathrm{mmHg}\).

What volume of ethylene glycol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH},\right.\) density \(=1.12 \mathrm{g} \mathrm{mL}^{-1}\) ) must be added to \(20.0 \mathrm{L}\) of water \(\left(K_{\mathrm{f}}=1.86^{\circ} \mathrm{C} / m\right)\) to produce a solution that freezes at \(-10^{\circ} \mathrm{C} ?\)

At \(25^{\circ} \mathrm{C}\) a \(0.50 \mathrm{g}\) sample of polyisobutylene (a polymer used in synthetic rubber) in \(100.0 \mathrm{mL}\) of benzene solution has an osmotic pressure that supports a \(5.1 \mathrm{mm}\) column of solution \((d=0.88 \mathrm{g} / \mathrm{mL}) .\) What is the molar mass of the polyisobutylene? (For \(\mathrm{Hg}\), \(d=13.6 \mathrm{g} / \mathrm{mL} .)\)

Nitrobenzene, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2},\) and benzene, \(\mathrm{C}_{6} \mathrm{H}_{6},\) are completely miscible in each other. Other properties of the two liquids are nitrobenzene: \(\mathrm{fp}=5.7^{\circ} \mathrm{C}, K_{\mathrm{f}}=\) \(8.1^{\circ} \mathrm{C} m^{-1} ;\) benzene: \(\mathrm{fp}=5.5^{\circ} \mathrm{C}, K_{\mathrm{f}}=5.12^{\circ} \mathrm{C} m^{-1} . \mathrm{It}\) is possible to prepare two different solutions with these two liquids having a freezing point of \(0.0^{\circ} \mathrm{C}\) What are the compositions of these two solutions, expressed as mass percent nitrobenzene?

The concentration of Ar in the ocean at \(25^{\circ} \mathrm{C}\) is \(11.5 \mu \mathrm{M} .\) The Henry's law constant for \(\mathrm{Ar}\) is \(1.5 \times 10^{-3}\) \(\mathrm{mol} \mathrm{L}^{-1} \mathrm{atm}^{-1} .\) Calculate the mass of \(\mathrm{Ar}\) in a liter of ocean water. Calculate the partial pressure of \(\mathrm{Ar}\) in the atmosphere.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Physical 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