The alcohol content of hard liquor is normally given in terms of the "proof," which is defined as twice the percentage by volume of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) present. Calculate the number of grams of alcohol present in \(1.00 \mathrm{~L}\) of 75 -proof gin. The density of ethanol is \(0.798 \mathrm{~g} / \mathrm{mL}\)

Short Answer

Expert verified
The number of grams of alcohol present in 1.00 L of 75-proof gin is 299.25 g.

Step by step solution

01

Determine the Volume Percentage of Alcohol

The proof of the liquor is given to be 75. The proof is twice the volume percentage of alcohol. So to get the volume percentage, divide the proof by 2. Hence, \( \frac{75}{2} = 37.5\% \). Therefore, the gin is 37.5% ethanol by volume.
02

Convert Percentage to Volume

Since the volume of the gin is specified to be 1.00 L (or 1000 ml, since 1 L = 1000 ml), the alcohol volume in ml is given by multiplying the total volume with the volume percent (expressed as a decimal). Hence, \(37.5\% \) of \( 1000 \, ml\) is \( 0.375 \times 1000 = 375 \, ml\). Therefore, the volume of alcohol in the gin is 375 ml.
03

Convert Volume to Mass

The final step is to convert the volume of the alcohol in ml to its mass in grams. We do this by multiplying by the density (given in grams/ml). Consequently, \(375 \, ml \times 0.798 \, g/ml = 299.25 \, g\). Therefore, the mass of alcohol present in 1.00 L of 75-proof gin is 299.25 g.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proof Definition Chemistry
Understanding alcohol proof is essential in chemistry and also in everyday contexts such as bartending and consumer purchasing. The term 'proof' dates back to 16th-century England, where it was a test to assure the alcohol content of spirits. In chemistry today, alcohol proof is a measure that defines the strength of the spirit, which is simply twice the alcohol by volume (ABV) percentage. For example, if a bottle of whisky is labeled as 75 proof, it means that it contains 37.5% ABV.

When calculating the alcohol content, as seen in the provided exercise, knowing the proof allows us to determine the volume percentage of ethanol in the solution. This knowledge is particularly useful not just in chemistry, but also in fields like pharmacology and the food industry, where precise alcohol measurements are necessary for formulations and regulations.
Converting Volume Percentage to Mass
When dealing with solutions, it's often necessary to convert volume percentages to mass, especially when precise quantitative data is needed. In chemistry, this conversion is fundamental for preparing solutions with specific concentration or assessing the quantity of a substance in a mixture.

The conversion process involves two main steps: firstly, determining the volume of the solvent or solute in the solution, and secondly, multiplying this volume by the density of the substance to obtain its mass. As illustrated in the exercise, after finding out the volume percentage of ethanol in the gin, multiplying it by the total volume provides us with the volume of alcohol. Then, applying the known density of ethanol (\(0.798 \text{g/mL}\)) to this volume yields the mass of alcohol in grams.

This conversion is critical in various scientific fields, including pharmacology, where dosing must be precise, in environmental science for pollution assessments, and in the culinary arts for recipe development.
Density of Ethanol
Density is a fundamental concept in science, defined as mass per unit volume, and it is crucial when we work with substances like ethanol (\(C_2H_5OH\)). The density of ethanol is particularly important in the fields of chemistry and biochemistry as it influences the behavior of ethanol in mixtures and reactions.

In the case of calculating the mass of ethanol in a given volume of alcoholic beverage—as in our example exercise—the known value of ethanol's density (\(0.798 \text{g/mL}\)) enables us to convert the volume of ethanol to its corresponding mass. This property of ethanol is exploited in various techniques, such as distillation, where the difference in densities between ethanol and water helps separate the two.

Moreover, understanding ethanol’s density is also imperative in fuel technology, where ethanol-blended fuels are assessed for their energy contents, and in the healthcare industry, where alcohol’s density is used for preparing sanitizers and disinfectants at precise concentrations.

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

Arrange the following aqueous solutions in order of decreasing freezing point, and explain your reasoning: \(0.50 \mathrm{~m} \mathrm{HCl}, 0.50 \mathrm{~m}\) glucose, \(0.50 \mathrm{~m}\) acetic acid.

As we know, if a soft drink can is shaken and then opened, the drink escapes violently. However, if after shaking the can we tap it several times with a metal spoon, no such "explosion" of the drink occurs. Why?

A very long pipe is capped at one end with a semipermeable membrane. How deep (in meters) must the pipe be immersed into the sea for freshwater to begin to pass through the membrane? Assume the water to be at \(20^{\circ} \mathrm{C}\) and treat it as a \(0.70 \mathrm{M} \mathrm{NaCl}\) solution. The density of seawater is \(1.03 \mathrm{~g} / \mathrm{cm}^{3}\) and the acceleration due to gravity is \(9.81 \mathrm{~m} / \mathrm{s}^{2}\)

Here is an after-dinner trick. With guests still sitting at the table, the host provided each of them with a glass of water containing an ice cube floating on top and a piece of string about \(2-3\) in. in length. He then asked them to find a way to lift the ice cube without touching it by hand or using any other objects such as a spoon or fork. Explain how this task can be accomplished. (Hint: The table had not been cleared so the salt and pepper shakers were still there.

A solution of \(2.50 \mathrm{~g}\) of a compound having the empirical formula \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{P}\) in \(25.0 \mathrm{~g}\) of benzene is observed to freeze at \(4.3^{\circ} \mathrm{C}\). Calculate the molar mass of the solute and its molecular formula.

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