Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 68

For a solution containing \(6.38 \%\) para-diclorobenzene by mass in benzene, the density of the solution as a function of temperature ( \(t\) ) in the temperature range 15 to \(65^{\circ} \mathrm{C}\) is given by the equation \(d(\mathrm{g} / \mathrm{mL})=1.5794-1.836 \times 10^{-3}(t-15)\) At what temperature will the solution have a density of \(1.543 \mathrm{g} / \mathrm{mL} ?\)

Short Answer

Expert verified
The solution will have a density of 1.543 g/mL at a temperature of \(35^{\circ} C\).

Step by step solution

01

Analyzing the given function

The function for density provided in the problem is \(d(\mathrm{g} / \mathrm{mL})=1.5794-1.836 \times 10^{-3}(t-15)\). This function describes the relationship between the temperature (\(t\)) and the density of solution (\(d)\). The target is to find the temperature when density is \(1.543 g/mL\).
02

Solve the equation for \(t\)

Substitute \(d\) in the equation with \(1.543 g/mL\) to find the temperature at this density: \(1.543 = 1.5794 - 1.836 \times 10^{-3}(t-15)\).
03

Isolate the term with \(t\)

The next step is to isolate the term with the variable \(t\). To do that, subtract \(1.5794\) from both sides of the equation. Now the equation looks like this: \(-0.0364 = - 1.836 \times 10^{-3}(t-15)\).
04

Simplify the equation

Further simplify the equation to find \(t\). By multiplying both sides of the equation by \(-1\), we get \(0.0364 = 1.836 \times 10^{-3}(t-15)\). Divide both sides of the equation by \(1.836 \times 10^{-3}\) to solve for \(t - 15\). The result is \(t - 15 = 0.0364 / 1.836 \times 10^{-3}\). Calculate the result to get \(t - 15 = 19.8255\).
05

Solve for \(t\)

Finally, solve for \(t\) by adding 15 to both sides of the equation. So, \(t = 15 + 19.8255\). Calculate the result to get the final temperature of \(t = 34.8255 ^{\circ} C\). The solution should be rounded to the nearest integer.
06

Round the final result

Round to the nearest integer. So, the final temperature is \(t = 35 ^{\circ} C\).

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

Briefly describe each of the following ideas: (a) SI base units; (b) significant figures; (c) natural law; (d) exponential notation.

Which is the greater mass, \(3257 \mathrm{mg}\) or \(0.000475 \mathrm{kg} ?\) Explain.

A solution contains \(10.05 \%\) sucrose (cane sugar) by mass. What mass of the solution, in grams, is needed for an application that requires \(1.00 \mathrm{kg}\) sucrose?

Appendix E describes a useful study aid known as concept mapping. Using the method presented in Appendix \(E,\) construct a concept map illustrating the different concepts presented in Sections \(1-2,\) \(1-3,\) and \(1-4\).

Blood alcohol content (BAC) is sometimes reported in weight-volume percent and, when it is, a BAC of \(0.10 \%\) corresponds to \(0.10 \mathrm{g}\) ethyl alcohol per \(100 \mathrm{mL}\) of blood. In many jurisdictions, a person is considered legally intoxicated if his or her BAC is 0.10\%. Suppose that a 68 kg person has a total blood volume of 5.4 L and breaks down ethyl alcohol at a rate of 10.0 grams per hour. \(^{*}\) How many 145 mL glasses of wine, consumed over three hours, will produce a BAC of \(0.10 \%\) in this 68 kg person? Assume the wine has a density of \(1.01 \mathrm{g} / \mathrm{mL}\) and is \(11.5 \%\) ethyl alcohol by mass. (* The rate at which ethyl alcohol is broken down varies dramatically from person to person. The value given here for the rate is a realistic, but not necessarily accurate, value.)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

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