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An 80-proof brandy is a \(40 . \%\) (v/v) ethanol solution. The "proof" is twice the percent concentration of alcohol in the beverage. How many milliliters of alcohol are present in \(750 \mathrm{~mL}\) of brandy?

Short Answer

Expert verified
300 mL

Step by step solution

01

Understand proof and concentration

Recognize that 'proof' is a measure of the alcoholic strength of a beverage. In the US, the proof value is twice the percentage of alcohol by volume. So, an 80-proof brandy corresponds to a 40% ethanol solution.
02

Convert the percentage to a decimal

Convert the percentage concentration of ethanol to a decimal by dividing by 100. Therefore, 40% becomes 0.40.
03

Calculate the volume of ethanol

To find the volume of ethanol in 750 mL of brandy, multiply the total volume by the decimal concentration. So, the calculation is: \( 750 \text{ mL} \times 0.40 = 300 \text{ mL} \)

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Key Concepts

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

Proof in Alcoholic Beverages
Understanding 'proof' is essential when studying alcoholic beverages. In the United States, the proof of a beverage is a measure of its alcohol content. It is defined as twice the percentage of alcohol by volume. Hence, if a beverage is labeled as 80-proof, it means it contains 40% alcohol by volume. This standardized system makes it easy to understand the alcohol content, regardless of the type of alcoholic drink.
Volume/Volume Percentage
Volume/volume percentage, often abbreviated as (v/v), is a way of expressing the concentration of a solution. It specifically describes the amount of solute (ethanol, in this case) in a specific volume of solution. For example, a 40% (v/v) ethanol solution means that there are 40 milliliters of ethanol in every 100 milliliters of the solution. This method is commonly used in chemistry and the beverage industry to represent the concentration of liquids mixed together.
Ethanol Concentration Calculations
Calculating the concentration of ethanol in a solution involves a few simple steps. To find out how much ethanol is present in a given volume of solution, you first need to know the percentage concentration. Let's revisit the example from the exercise: an 80-proof brandy has a 40% (v/v) ethanol concentration. To convert this percentage to a usable form, you must convert it to a decimal by dividing by 100 (i.e., 40% becomes 0.40). Using this decimal, you simply multiply the total volume of the solution by the ethanol concentration. So for 750 mL of brandy, the calculation would be:
\[ 750 \text{ mL} \times 0.40 = 300 \text{ mL} \]
This means there are 300 mL of ethanol in 750 mL of brandy.
Conversion of Percentage to Decimal
Converting a percentage to a decimal is straightforward but crucial for accurate calculations. To perform this conversion, you divide the percentage value by 100. This operation removes the percentage sign and converts the number into a form that is easy to use in multiplication and other mathematical operations. For example, converting 40% to a decimal involves these steps:
  • Start with the percentage: 40%
  • Divide by 100: 40/100 = 0.40

Now, you can easily use the decimal in your calculations, such as finding the volume of ethanol in a solution.

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Most popular questions from this chapter

A mouthwash contains \(22.5 \%(\mathrm{v} / \mathrm{v})\) alcohol. If the bottle of mouthwash contains \(355 \mathrm{~mL}\), what is the volume, in milliliters, of alcohol?

Calculate the concentration of each of the following diluted solutions: a. \(1.0 \mathrm{~L}\) of a \(4.0 \mathrm{M} \mathrm{HNO}_{3}\) solution is added to water so that the final volume is \(8.0 \mathrm{~L}\). b. Water is added to \(0.25 \mathrm{~L}\) of a \(6.0 \mathrm{M}\) NaF solution to make 2.0 L of a diluted NaF solution. c. A \(50.0-\mathrm{mL}\) sample of an \(8.0 \%(\mathrm{~m} / \mathrm{v}) \mathrm{KBr}\) solution is diluted with water so that the final volume is \(200.0 \mathrm{~mL}\). d. A \(5.0\) -mL sample of a \(50.0 \%\) (m/v) acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) solution is added to water to give a final volume of \(25 \mathrm{~mL}\).

How many milliliters of each of the following solutions will provide \(25.0 \mathrm{~g}\) of \(\mathrm{KOH}\) ? a. \(2.50 \mathrm{M} \mathrm{KOH}\) solution b. \(0.750 \mathrm{M}\) KOH solution c. \(5.60 \mathrm{M} \mathrm{KOH}\) solution

Calculate the grams of solute needed to prepare each of the following solutions: a. \(2.00 \mathrm{~L}\) of a \(1.50 \mathrm{M} \mathrm{NaOH}\) solution b. \(4.00 \mathrm{~L}\) of a \(0.200 \mathrm{M} \mathrm{KCl}\) solution c. \(25.0 \mathrm{~mL}\) of a \(6.00 \mathrm{M} \mathrm{HCl}\) solution

Calculate the molarity of the solution when water is added to prepare each of the following: a. \(25.0 \mathrm{~mL}\) of a \(0.200 \mathrm{M}\) NaBr solution is diluted to \(50.0 \mathrm{~mL}\) b. \(15.0 \mathrm{~mL}\) of a \(1.20 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4}\) solution is diluted to \(40.0 \mathrm{~mL}\) c. \(75.0 \mathrm{~mL}\) of a \(6.00 \mathrm{M} \mathrm{NaOH}\) solution is diluted to \(255 \mathrm{~mL}\)

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