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Chapter 14: Problem 71

You love the smell of jasmine and have bought a solution of methyl jasmonate, one of the compounds responsible for the characteristic odor of jasmine. If the solution is composed of a \(0.5298\)-molar solution of methyl jasmonate in acetone, how many molecules of methyl jasmonate will you spread around the room if you spritz \(0.414 \mathrm{~mL}\) of the solution into the air?

Short Answer

Expert verified
1.3214 x 10^{20} molecules

Step by step solution

01

- Calculate the number of moles of methyl jasmonate

First, convert the volume of the solution from milliliters to liters. Given volume = 0.414 mL. Convert mL to liters: \[ 0.414 \text{ mL} = 0.414 \times 10^{-3} \text{ L} = 4.14 \times 10^{-4} \text{ L} \]Now, use the molarity (M) to find the number of moles. Number of moles = Molarity (M) * Volume (L) \[ \text{Molarity} = 0.5298 \text{ M} \]Thus, \[ \text{Number of moles} = 0.5298 \text{ M} \times 4.14 \times 10^{-4} \text{ L} = 2.194 \times 10^{-4} \text{ moles} \]
02

- Calculate the number of molecules of methyl jasmonate

Use Avogadro's number (\[ 6.022 \times 10^{23} \text{ molecules/mole} \] ) to convert moles to molecules. Number of molecules = Number of moles * Avogadro's number Thus, \[ \text{Number of molecules} = 2.194 \times 10^{-4} \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mole} \]\[ = 1.3214 \times 10^{20} \text{ molecules} \]

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Key Concepts

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

Molarity Calculations
Molarity is a measure of concentration, indicating how many moles of a solute are present in one liter of solution. To calculate molarity, we use the formula:

\( \text{Molarity} (M) = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \)

In the problem, we are given a solution with a molarity of 0.5298 M and a volume of 0.414 mL, which we need to convert to liters. Since 1 mL = 0.001 L, we convert the volume by multiplying by \(10^{-3}\).

For instance:
  • Volume in liters: 0.414 mL * \(10^{-3}\) = 0.000414 L = \(4.14 \times 10^{-4}\text{ L}\)
With the volume in liters, we can find the number of moles using the molarity:
  • Number of moles = Molarity * Volume in liters
  • Number of moles = 0.5298 M * \(4.14 \times 10^{-4}\text {L}\) = \(2.194 \times 10^{-4}\text{ moles}\)
Molarity calculations are essential for solving problems related to solution concentration.
Avogadro's Number
Avogadro's Number is a key concept in chemistry. It tells us the number of particles, atoms, or molecules in one mole of a substance. Avogadro's Number is:

\( 6.022 \times 10^{23} \text{ particles/mol} \)

It helps convert between the number of moles and the number of particles. In the exercise, we first found the number of moles of methyl jasmonate to be \(2.194 \times 10^{-4} \text{ moles}\).
  • Using Avogadro's Number, we can find the number of molecules:
    Number of molecules = Number of moles * Avogadro's Number
So, we get:
  • Number of molecules = \(2.194 \times 10^{-4} \text{ moles} \) * \( 6.022 \times 10^{23} \text{ molecules/mole}\)
Therefore, the number of molecules = \(1.3214 \times 10^{20} \text{ molecules}\).
Understanding Avogadro's Number is important for converting moles to the actual number of molecules or atoms.
Concentration Conversions
Concentration conversions are frequently needed in chemistry to change the units of concentration or to use these units in calculations. One common conversion is converting volume measurements. In the example problem, we converted milliliters to liters to use them in molarity calculations. Here is why and how:

  • Given volume in milliliters: 0.414 mL
  • Because 1 mL = 0.001 L, we multiply the volume in mL by \(10^{-3}\) to convert it to liters.
Thus:

\(0.414\text{ mL} = 0.414 \times 10^{-3} \text{ L} = 4.14 \times 10^{-4} \text{ L}\)

Converting volumes to liters allows us to easily use the molarity formula.
  • Molarity calculations are usually done with volume in liters to keep units consistent.
  • Accurate conversions ensure correct concentration calculations.

Understanding the process of concentration conversions is crucial for accurate chemical calculations and real-world applications.

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Most popular questions from this chapter

If \(150 \mathrm{~mL}\) of \(0.055 \mathrm{M} \mathrm{HNO}_{3}\) are needed to completely neutralize \(1.48 \mathrm{~g}\) of an impure sample of sodium hydrogen carbonate (baking soda), what percent of the sample is baking soda?

What happens to sugar molecules \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) when they are dissolved in water?

Law enforcement uses a quick and easy test for the presence of the illicit drug \(\mathrm{PCP}\), reacting it with potassium iodide. The PCP will form a crystalline solid with a long branching needlelike structure with \(\mathrm{Kl}\). What is the molarity of \(\mathrm{Kl}\) in a stock solution prepared by dissolving \(396.1 \mathrm{~g}\) of \(\mathrm{Kl}\) to a total volume of \(750.0 \mathrm{~mL}\) ?

What is the molarity of a phosphoric acid solution if the solution is \(85 \%\) by mass \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and has a density of \(1.7 \mathrm{~g} / \mathrm{mL}\) ?

Calculate the molality of each of the following solutions: (a) \(2.0 \mathrm{~mol} \mathrm{HCl}\) in \(175 \mathrm{~g}\) water (b) \(14.5 \mathrm{~g} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) in \(550.0 \mathrm{~g}\) water (c) \(25.2 \mathrm{~mL}\) methanol, \(\mathrm{CH}_{3} \mathrm{OH}(d=0.791 \mathrm{~g} / \mathrm{mL})\) in \(595 \mathrm{~g}\) ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)
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