The term "proof," still used to describe the ethanol content of alcoholic beverages, originated in seventeenthcentury England. A sample of whiskey was poured on gunpowder and set afire. If the gunpowder ignited after the whiskey had burned off, this "proved" that the whiskey had not been watered down. The minimum ethanol content for a positive test was about \(50 \%\) by volume. The \(50 \%\) ethanol solution became known as \(^{\prime \prime} 100\) proof." Thus, an 80 -proof whiskey would be \(40 \% \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) by volume. Listed in the table below are some data for several aqueous solutions of ethanol. With a minimum amount ofcalculation, determine which of the solutions are more than 100 proof. Assume that the density of pure ethanol is \(0.79 \mathrm{g} / \mathrm{mL}\).$$\begin{array}{cl} \hline \text { Molarity of Ethanol, } M & \text { Density of Solution, g/mL } \\\ \hline 4.00 & 0.970 \\\5.00 & 0.963 \\\6.00 & 0.955 \\\7.00 & 0.947 \\\8.00 & 0.936 \\\9.00 & 0.926 \\\10.00 & 0.913 \\\\\hline\end{array}.$$

Short Answer

Expert verified
After calculating the volume percentage for each solution using the given molarity and solution density, solutions with a percentage greater than 50% are more than 100 proof.

Step by step solution

01

Understand proof

The term 'Proof' is used to describe the ethanol content of alcoholic beverages. A 100-proof whiskey has 50% alcohol by volume. Therefore, to determine if a solution is more than 100 proof, we need to calculate if its alcohol content is more than 50% by volume.
02

Calculate the volume percent of ethanol

The volume percentage of ethanol can be calculated from its molarity and the density of the solution using the formula: \(V \% = \frac{(Molarity \times Molar mass \times 1000)}{(Density \times 10)}\% where Molar mass of ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\), is 46.07 g/mol. Assume the density of pure ethanol is 0.79 g/mL. The 4.00 M solution, for example, would be calculated as: \(V \% = \frac{(4.00 mol/L \times 46.07 g/mol) \times 1000}{0.970 g/mL \times 10}\% \)
03

Compare with 100 proof

Given that 100 proof equals 50% ethanol by volume, any solution with a percent volume greater than 50% would be more than 100 proof. Apply the proof determination to all solution densities.
04

Repeat steps 2 and 3

Repeat steps 2 and 3 for the remaining molarities and compare their volume percents to determine if they are more than 100 proof.

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 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

Calculate the mole fraction of the solute in the following aqueous solutions:(a) \(0.112 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) \((d=1.006 \mathrm{g} / \mathrm{mL})\) (b) \(3.20 \%\) ethanol,by volume \((d=0.993 \mathrm{g} / \mathrm{mL}\) pure \(\left.\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}, d=0.789 \mathrm{g} / \mathrm{mL}\right)\).

A solution is prepared by mixing \(1.28 \mathrm{mol} \mathrm{C}_{7} \mathrm{H}_{16}\) \(2.92 \mathrm{mol} \mathrm{C}_{8} \mathrm{H}_{18},\) and \(2.64 \mathrm{mol} \mathrm{C}_{9} \mathrm{H}_{20} .\) What is the (a) mole fraction and (b) mole percent of each component of the solution?

An isotonic solution is described as \(0.92 \%\) NaCl (mass/volume). Would this also be the required concentration for isotonic solutions of other salts, such as \(\mathrm{KCl}, \mathrm{MgCl}_{2},\) or \(\mathrm{MgSO}_{4} ?\) Explain.

The aqueous solubility at \(20^{\circ} \mathrm{C}\) of \(\mathrm{Ar}\) at \(1.00 \mathrm{atm}\) is equivalent to \(33.7 \mathrm{mL} \mathrm{Ar}(\mathrm{g}),\) measured at STP, per liter of water. What is the molarity of Ar in water that is saturated with air at 1.00 atm and \(20^{\circ} \mathrm{C}\) ? Air contains \(0.934 \%\) Ar by volume. Assume that the volume of water does not change when it becomes saturated with air.

An aqueous solution has \(109.2 \mathrm{g} \mathrm{KOH} / \mathrm{L}\) solution. The solution density is \(1.09 \mathrm{g} / \mathrm{mL} .\) Your task is to use \(100.0 \mathrm{mL}\) of this solution to prepare \(0.250 \mathrm{m}\) KOH. What mass of which component, \(\mathrm{KOH}\) or \(\mathrm{H}_{2} \mathrm{O}\), would you add to the \(100.0 \mathrm{mL}\) of solution?

See all solutions

Recommended explanations on Chemistry Textbooks

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