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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 with a 6.86 M concentration, you need to dissolve 315.88 grams of ethanol in water. (b) To make 1.00 L of vodka, you need to add 400.36 mL of ethanol.

Step by step solution

01

1. Determine the moles of ethanol needed in the solution

First, we need to find the moles of ethanol we require to make 1.00 L of vodka with a 6.86 M concentration. The formula for molarity (M) is: M = moles of solute / volume of solution (in liters) We are given the molarity and volume, so we can solve for moles: moles = M 脳 volume moles = 6.86 M 脳 1.00 L moles = 6.86 mol
02

2. Calculate the grams of ethanol needed

Now we need to convert the moles of ethanol into grams. The molar mass of ethanol can be found by adding the individual molar masses of its elements: Ethanol: \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}\) Carbon: 12.01 g/mol (2 atoms) Hydrogen: 1.01 g/mol (6 atoms) Oxygen: 16.00 g/mol (1 atom) Molar mass of ethanol = (2 脳 12.01) + (6 脳 1.01) + (1 脳 16.00) = 46.07 g/mol To find the grams of ethanol we need, we'll multiply the moles of ethanol by its molar mass: grams of ethanol = moles 脳 molar mass grams of ethanol = 6.86 mol 脳 46.07 g/mol
03

3. Calculate the volume of ethanol required

We are given the density of ethanol (0.789 g/mL) and are asked to calculate the volume in milliliters. To find the volume, we'll use the formula: volume = mass / density volume = grams of ethanol / (0.789 g/mL) The final output will be our answers for parts (a) and (b). Solution:
04

1. Determine the moles of ethanol needed in the solution

moles = 6.86 M 脳 1.00 L moles = 6.86 mol
05

2. Calculate the grams of ethanol needed

Molar mass of ethanol = 46.07 g/mol grams of ethanol = 6.86 mol 脳 46.07 g/mol grams of ethanol = 315.88 g So, to make 1.00 L of vodka with a 6.86 M concentration, you need to dissolve 315.88 grams of ethanol in water.
06

3. Calculate the volume of ethanol required

volume = 315.88 g / (0.789 g/mL) volume = 400.36 mL So, to make 1.00 L of vodka, you need to add 400.36 mL of ethanol.

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Key Concepts

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

Ethanol Solutions
Ethanol, also known as ethyl alcohol, is a key component in many solutions, particularly in alcoholic beverages like vodka. When preparing an ethanol solution, the goal is to achieve a desired concentration, usually measured in molarity (M), which reflects the number of moles of solute per liter of solution. In our original exercise, we are tasked with preparing a solution with a specific molarity of ethanol.
  • An ethanol solution is formed by dissolving ethanol in water.
  • The solubility of ethanol in water allows it to easily mix, enabling the creation of solutions with various concentrations.
  • These solutions are held to desired specifications for both scientific experiments and culinary creations.
It鈥檚 important to understand the solution preparation process when working with ethanol, as incorrect measurements can lead to variations in concentration that may affect the solution鈥檚 intended use.
Molarity Calculations
Molarity is a crucial concept in chemistry that denotes the concentration of a solution. It's represented by the formula: \[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}}. \]In the provided exercise, the molarity of ethanol is 6.86 M in a 1.00 L solution. This means 6.86 moles of ethanol are needed. Here鈥檚 how it works:
  • First, identify the molarity required for the solution.
  • Multiply the molarity by the volume of the solution to calculate the amount of solute needed in moles.
  • This calculation ensures precise concentrations are achieved, essential for scientific accuracy.
For ethanol, converting moles to grams requires knowledge of its molar mass, which is 46.07 g/mol. This conversion is done through multiplying the number of moles by the molar mass, evening out the concentration in terms that are practical for measuring and mixing.
Density Calculations
Density plays a significant role when determining how much volume a certain mass of substance will occupy. The formula for density is:\[ \text{density} = \frac{\text{mass}}{\text{volume}}. \]Given ethanol鈥檚 density of 0.789 g/mL, we can calculate the volume needed to reach a specific mass, useful for solution preparations and handling.To find the volume of ethanol required:
  • Start with the mass of ethanol required for the solution.
  • Use the formula \( \text{volume} = \frac{\text{mass}}{\text{density}} \) to find the equivalent volume in milliliters.
  • This calculation allows you to measure the correct amount of liquid ethanol to be used, ensuring the preparation's consistency.
Understanding density helps in accurately measuring liquids in laboratory settings and commercial distillation processes, ensuring the final product complies with specific standards.

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Most popular questions from this chapter

The metal cadmium tends to form \(\mathrm{Cd}^{2+}\) ions. The following observations are made: (i) When a strip of zinc metal is placed in \(\mathrm{CdCl}_{2}(a q),\) cadmium metal is deposited on the strip. (ii) When a strip of cadmium metal is placed in \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q),\) nickel metal is deposited on the strip. (a) Write net ionic equations to explain each of the preceding observations. (b) Which elements more closely define the position of cadmium in the activity series? (c) What experiments would you need to perform to locate more precisely the position of cadmium in the activity series?

Indicate the concentration of each ion or molecule present in the following solutions: (a) 0.25\(M\) NaNO \(_{3}\) , (b) \(1.3 \times 10^{-2} M \mathrm{MgSO}_{4},(\mathbf{c}) 0.0150 M \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},(\mathbf{d})\) a mixture of 45.0 \(\mathrm{mL}\) of 0.272 \(\mathrm{M} \mathrm{NaCl}\) and 65.0 \(\mathrm{mL}\) of 0.0247 \(\mathrm{M}\) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} .\) Assume that the volumes are additive.

Write balanced molecular and net ionic equations for the reactions of (a) manganese with dilute sulfuric acid, (b) chromium with hydrobromic acid, (c) tin with hydrochloric acid, (d) aluminum with formic acid, HCOOH.

The concept of chemical equilibrium is very important. Which one of the following statements is the most correct way to think about equilibrium? (a) If a system is at equilibrium, nothing is happening. (b) If a system is at equilibrium, the rate of the forward reaction is equal to the rate of the back reaction. (c) If a system is at equilibrium, the product concentration is changing over time. Section 4.1\(]\)

You want to analyze a silver nitrate solution. (a) You could add \(\mathrm{HCl}(a q)\) to the solution to precipitate out AgCl(s). What volume of a 0.150 \(\mathrm{M} \mathrm{HCl}(a q)\) solution is needed to precipitate the silver ions from 15.0 \(\mathrm{mL}\) of a 0.200 \(\mathrm{MgNO}_{3}\) solution? (b) You could add solid \(\mathrm{KCl}\) to the solution to precipitate out AgCl(s). What mass of KCl is needed to precipitate the silver ions from 15.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M} \mathrm{AgNO}_{3}\) solution? (c) Given that a 0.150 \(\mathrm{M} \mathrm{HCl}(a q)\) solution costs \(\$ 39.95\) for 500 \(\mathrm{mL}\) and that \(\mathrm{KCl}\) costs \(\$ 10 / \mathrm{ton},\) which analysis procedure is more cost-effective?
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