Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 144

What should be the density of an aqueous solution of urea (molar mass \(=60 \mathrm{~g} / \mathrm{mol}\) ) such that the molality and molarity of the solution become equal? (a) \(1.0 \mathrm{~g} / \mathrm{ml}\) (b) \(1.6 \mathrm{~g} / \mathrm{ml}\) (c) \(1.06 \mathrm{~g} / \mathrm{m}]\) (d) \(1.16 \mathrm{~g} / \mathrm{m}]\)

Short Answer

Expert verified
Answer: (b) 1.6 g/ml

Step by step solution

01

Understand molality and molarity formulas

Molality (m) is defined as moles of solute per kilogram of solvent whereas Molarity (M) is defined as moles of solute per liter of the solution. Molality (m) formula is: m = moles of urea / mass of water (in kg) Molarity (M) formula is: M = moles of urea / volume of the solution (in L)
02

Establish equations for molality and molarity

Let x be the moles of urea. Molality (m) = x moles / (1000 - 60x) g, since mass of water = (1000 - 60x) g, which is the mass of 1 L of the solution (1000 g) minus the mass of urea (60x g) Molarity (M) = x moles / (density × (1000 - 60x) / 1000) L, since volume of the solution = mass of solution/density and mass of solution = 1000g According to the problem, the molality and molarity of the solution are equal. Therefore, m = M x moles / (1000 - 60x) g = x moles / (density × (1000 - 60x) / 1000) L
03

Simplify and calculate the density

Cancel the 'x moles' from both sides and rearrange the equation to isolate the density: density = 1000 / (1000 - 60x) Now, we will substitute the given density options to find which one makes the equation true.
04

Test the provided density options

(a) \(1.0 \mathrm{~g} / \mathrm{ml}\) (b) \(1.6 \mathrm{~g} / \mathrm{ml}\) (c) \(1.06 \mathrm{~g} / \mathrm{m]}\) (d) \(1.16 \mathrm{~g} / \mathrm{m]}\) (a) If the density is \(1.0 \mathrm{~g}/\mathrm{ml}\), we get: 1 = 1000 / (1000 - 60x) x = 0 In this case, the number of moles of urea would be zero, which is not possible since a solution must have urea in it to have a molality and molarity. (b) If the density is \(1.6 \mathrm{~g}/\mathrm{ml}\), we get: 1.6 = 1000 / (1000 - 60x) x = 0.6 In this case, we have a valid non-zero value of x. (c) and (d) are incorrect units for density, so they cannot be the answer.
05

Conclusion

Based on the calculations and analysis, the density of the aqueous urea solution that makes the molality and molarity equal is: (b) \(1.6 \mathrm{~g} / \mathrm{ml}\)

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 density (in \(\mathrm{g} /\) l) of an equimolar mixture of methane and ethane at 1 atm and \(0^{\circ} \mathrm{C}\) is (a) \(1.03\) (b) \(2.05\) (c) \(0.94\) (d) \(1.25\)

A sample of ozone gas is found to be \(40 \%\) dissociated into oxygen. The average molecular mass of sample should be

The minimum mass of water needed to slake \(1 \mathrm{~kg}\) of quicklime, assuming no loss by evaporation, is (a) \(243.2 \mathrm{~g}\) (b) \(642.8 \mathrm{~g}\) (c) \(160.7 \mathrm{~g}\) (d) \(321.4 \mathrm{~g}\)

At room temperature, the molar volume of hydrogen fluoride gas has a mass of about \(50 \mathrm{~g}\). The formula weight of hydrogen fluoride is 20 . Gaseous hydrogen fluoride at room temperature is therefore, probably a mixture of (a) \(\mathrm{H}_{2}\) and \(\mathrm{F}_{2}\) (b) \(\mathrm{HF}\) and \(\mathrm{H}_{2} \mathrm{~F}_{2}\) (c) \(\mathrm{HF}\) and \(\mathrm{H}_{2, s} \mathrm{~F}_{2.5}\) (d) \(\mathrm{H}_{2} \mathrm{~F}_{2}\) and \(\mathrm{H}_{3} \mathrm{~F}_{3}\)

Hydrogen cyanide, HCN, is prepared from ammonia, air and natural gas \(\left(\mathrm{CH}_{4}\right)\) by the following process. \(2 \mathrm{NH}_{3}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{CH}_{4}(\mathrm{~g}) \stackrel{\mathrm{Pt}}{\longrightarrow}\) \(2 \mathrm{HCN}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) If a reaction vessel contains \(11.5 \mathrm{~g} \mathrm{NH}_{3}\), \(10.0 \mathrm{~g} \mathrm{O}_{2}\), and \(10.5 \mathrm{~g} \mathrm{CH}_{4}\), what is the maximum mass, in grams, of hydrogen cyanide that could be made, assuming the reaction goes to completion? (a) \(18.26 \mathrm{~g}\) (b) \(5.625 \mathrm{~g}\) (c) \(17.72 \mathrm{~g}\) (d) \(16.875 \mathrm{~g}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Nuclear 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