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Chapter 10: Problem 96

The term proof is defined as twice the percent by volume of pure ethanol in solution. Thus, a solution that is \(95 \%\) (by volume) ethanol is 190 proof. What is the molarity of ethanol in a 92 proof ethanol-water solution? Assume the density of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH},\) is \(0.79 \mathrm{g} / \mathrm{cm}^{3}\) and the density of water is \(1.0 \mathrm{g} / \mathrm{cm}^{3}\)

Short Answer

Expert verified
The molarity of ethanol in a 92 proof ethanol-water solution is approximately \(7.88\, M\).

Step by step solution

01

Calculate the volume percentage of ethanol in the solution

The problem states that the proof of a solution is defined as twice the percent by volume of pure ethanol in the solution. Given that the solution we are working with is 92 proof, we will find the volume percentage of ethanol in the solution by dividing 92 by 2: \[Volume \, Percentage \, of \, Ethanol = \frac{92}{2} = 46 \%\]
02

Use the densities to calculate mass percentages

We are given the density of ethanol (\(0.79 \, g/cm^3\)) and the density of water (\(1.0 \, g/cm^3\)). We can use these densities to find the mass percentages of ethanol and water in the solution: \[Mass \, Percentage \, of \, Ethanol = 0.46 * 0.79\] \[Mass \, Percentage \, of \, Water = 0.54 * 1.0\]
03

Convert mass percentages to actual masses

Assume we have 100 mL of the solution. We can find the mass of ethanol and water in the 100 mL solution: \[Mass \, of \, Ethanol = 100 * 0.46 * 0.79 = 36.34 \, g\] \[Mass \, of \, Water = 100 * 0.54 * 1.0 = 54 \, g\]
04

Convert the mass of ethanol to moles

The molar mass of ethanol is approximately \(46.07\, g/mol\). We can use this to convert the mass of ethanol into moles: \[Moles \, of \, Ethanol = \frac{36.34}{46.07} = 0.788 \, moles\]
05

Calculate the volume of the solution

We have assumed 100 mL of the solution and therefore the total volume is equal to 100 mL. Now, convert the volume from mL to L: \[Total \, Volume = 100 \, mL * \frac{1 \, L}{1000 \, mL} = 0.1 \, L\]
06

Determine the molarity of ethanol in the solution

Now, we can determine the molarity of ethanol in the solution: \[Molarity \, of \, Ethanol = \frac{Moles \, of \, Ethanol}{Total \, Volume} = \frac{0.788}{0.1} = 7.88 \, M\] The molarity of ethanol in the 92 proof ethanol-water solution is approximately \(7.88\, M\).

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Key Concepts

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

Proof
When we talk about 'proof' in the context of ethanol solutions, it's a term often used in alcoholic beverage labeling to describe alcohol content. The proof number is twice the percentage of ethanol by volume present in the solution.
  • If a solution is 100 proof, that means it contains 50% ethanol by volume.
  • Understanding proof helps determine the strength of alcoholic drinks, and it's crucial in solution chemistry calculations.
For instance, in a 92 proof ethanol solution, the ethanol volume percentage is 46%. Knowing this percentage helps us calculate other important figures like molarity.
Ethanol
Ethanol, often known by its chemical formula \( ext{C}_2 ext{H}_5 ext{OH}\), is a volatile, flammable, and colorless liquid. It's the type of alcohol found in alcoholic beverages, used as a solvent, and also has medical and fuel applications. This organic compound plays a central role in solution chemistry and particularly in preparing alcohol solutions.
  • Ethanol's structure: A simple molecule with one oxygen atom and a two-carbon chain.
  • Key properties: Boiling point of about 78.37掳C, making it volatile.
  • Mixes well with water due to its polar hydroxyl group (-OH).
Understanding ethanol is crucial for figuring out its interactions in solutions.
Density of Ethanol
The density of ethanol is critical for calculating its concentration in a solution. It's an essential property that tells us the mass of ethanol contained in a certain volume. In this exercise, ethanol is given a density of \(0.79 \, ext{g/cm}^3\).
  • This means that one cubic centimeter (or milliliter) of ethanol weighs 0.79 grams.
  • Density can change slightly with temperature and pressure but is often considered constant for calculations at room temperature.
  • Knowing the density allows us to convert volume to mass, which is especially handy when determining the molarity of ethanol in a mixture.
Density acts as a bridge between volume and mass, crucial for accurate solution chemistry computations.
Solution Chemistry
Solution chemistry involves studying how substances dissolve, creating a homogeneous solution. It's about understanding the interactions at play when a solute (like ethanol) is mixed with a solvent (like water).
  • Concentration measures such as molarity are key focus areas. Molarity refers to the number of moles of solute per liter of solution.
  • When calculating molarity, we often use conversion steps that involve the compound's density and molar mass.
  • In our example, we determined that the molarity of ethanol in a 92 proof, which is 46% ethanol by volume, is around \(7.88 \, ext{M}\).
Mastering solution chemistry helps in formulating mixtures accurately for academic, industrial, and practical applications.

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Most popular questions from this chapter

How would you prepare 1.0 L of an aqueous solution of sodium chloride having an osmotic pressure of 15 atm at \(22^{\circ} \mathrm{C} ?\) Assume sodium chloride exists as \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions in solution.

A water desalination plant is set up near a salt marsh containing water that is 0.10 \(M\) NaCl. Calculate the minimum pressure that must be applied at \(20 .^{\circ} \mathrm{C}\) to purify the water by reverse osmosis. Assume \(\mathrm{NaCl}\) is completely dissociated.

Is molality or molarity dependent on temperature? Explain your answer. Why is molality, and not molarity, used in the equations describing freezing-point depression and boiling point elevation?

The vapor pressure of a solution containing 53.6 g glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right)\) in \(133.7 \mathrm{g}\) ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is 113 torr at \(40^{\circ} \mathrm{C}\) Calculate the vapor pressure of pure ethanol at \(40^{\circ} \mathrm{C}\) assuming that glycerin is a nonvolatile, nonelectrolyte solute in ethanol.

The solubility of nitrogen in water is \(8.21 \times 10^{-4} \mathrm{mol} / \mathrm{L}\) at \(0^{\circ} \mathrm{C}\) when the \(\mathrm{N}_{2}\) pressure above water is 0.790 atm. Calculate the Henry's law constant for \(\mathrm{N}_{2}\) in units of mol/L \cdot atm for Henry's law in the form \(C=k P,\) where \(C\) is the gas concentration in mol/L. Calculate the solubility of \(\mathrm{N}_{2}\) in water when the partial pressure of nitrogen above water is 1.10 atm at \(0^{\circ} \mathrm{C}\).
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