/*! 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 67 (a) How many grams of ethanol, \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 67

(a) How many grams of ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\), should you dissolve in water to make \(1.00 \mathrm{~L}\) of vodka (which is an aqueous solution that is \(6.86 \mathrm{M}\) ethanol)? (b) Using the density of ethanol \((0.789 \mathrm{~g} / \mathrm{mL})\), calculate the volume of ethanol you need to make \(1.00 \mathrm{~L}\) of vodka.

Short Answer

Expert verified
(a) To make 1.00 L of vodka, you need to dissolve 315.92 grams of ethanol, CH3CH2OH, in water. (b) Using the density of ethanol (0.789 g/mL), you need 400.51 mL of ethanol to make 1.00 L of vodka.

Step by step solution

01

Identify the given information and desired outcomes

We are given: (a) Molarity (M) of ethanol in vodka: 6.86 M (b) Density of ethanol: 0.789 g/mL We need to find: (a) Grams of ethanol required to make 1 L of vodka (b) Volume of ethanol required to make 1 L of vodka
02

Calculation of moles of ethanol needed

Since we know the molarity and volume of the solution, we can calculate the moles of ethanol required. Molarity is defined as moles per liter, so we can multiply the volume with the molarity to calculate the total moles of ethanol needed: Moles of ethanol = Molarity 脳 Volume Moles of ethanol = 6.86 mol/L 脳 1 L Moles of ethanol = 6.86 mol
03

Calculation of grams of ethanol needed

Now that we have calculated the moles of ethanol needed, we can convert this to grams of ethanol using the molecular weight of ethanol, which is: Molecular weight of ethanol, CH3CH2OH = 12.01 g/mol (C) + 1.01 g/mol (6H) + 16.00 g/mol (O) Molecular weight of ethanol = 46.07 g/mol Using the moles of ethanol and molecular weight, we can calculate the grams of ethanol: Grams of ethanol = Moles 脳 Molecular weight Grams of ethanol = 6.86 mol 脳 46.07 g/mol Grams of ethanol = 315.92 g Therefore, 315.92 grams of ethanol is required to make 1 L of vodka.
04

Calculation of the volume of ethanol needed

Now that we know the grams of ethanol required, we can calculate the volume using the given density: Volume of ethanol = Mass / Density Volume of ethanol = 315.92 g / (0.789 g/mL) Volume of ethanol = 400.51 mL Therefore, 400.51 mL of ethanol is required to make 1 L of vodka.

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

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.

(a) By titration, \(15.0 \mathrm{~mL}\) of \(0.1008 \mathrm{M}\) sodium hydroxide is needed to neutralize a \(0.2053-\mathrm{g}\) sample of a weak acid. What is the molar mass of the acid if it is monoprotic? (b) An elemental analysis of the acid indicates that it is composed of \(5.89 \% \mathrm{H}, 70.6 \% \mathrm{C}\), and \(23.5 \% \mathrm{O}\) by mass. What is its molecular formula?

The arsenic in a \(1.22-\mathrm{g}\) sample of a pesticide was converted to \(\mathrm{AsO}_{4}{ }^{3-}\) by suitable chemical treatment. It was then titrated using \(\mathrm{Ag}^{+}\)to form \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) as a precipitate. (a) What is the oxidation state of \(\mathrm{As}\) in \(\mathrm{AsO}_{4}{ }^{3-}\) ? (b) \(\mathrm{Name} \mathrm{Ag}_{3} \mathrm{AsO}_{4}\) by analogy to the corresponding compound containing phosphorus in place of arsenic. (c) If it took \(25.0 \mathrm{~mL}\) of \(0.102 \mathrm{M} \mathrm{Ag}^{+}\)to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

The average concentration of gold in seawater is \(100 \mathrm{fM}\) (femtomolar). Given that the price of gold is \(\$ 1764.20\) per troy ounce ( 1 troy ounce \(=31.103 \mathrm{~g}\) ), how many liters of seawater would you need to process to collect \(\$ 5000\) worth of gold, assuming your processing technique captures only \(50 \%\) of the gold present in the samples?

The labels have fallen off three bottles containing powdered samples of metals; one contains zinc, one lead, and the other platinum. You have three solutions at your disposal: \(1 \mathrm{M}\) sodium nitrate, \(1 \mathrm{M}\) nitric acid, and \(1 \mathrm{M}\) nickel nitrate. How could you use these solutions to determine the identities of each metal powder? [Section 4.4]
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

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.