Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The allowable concentration level of vinyl chloride, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl},\) in the atmosphere in a chemical plant is \(2.0 \times 10^{-6} \mathrm{~g} / \mathrm{L}\). How many moles of vinyl chloride in each liter does this represent? How many molecules per liter?

Short Answer

Expert verified
The allowable concentration level of vinyl chloride, \(\mathrm{C}_{2}\mathrm{H}_{3} \mathrm{Cl}\), in the atmosphere represents approximately \(3.20 \times 10^{-8}\) moles per liter and approximately \(1.93 \times 10^{16}\) molecules per liter.

Step by step solution

01

Find the molecular weight of vinyl chloride

We are given the formula for vinyl chloride as \(\mathrm{C}_{2}\mathrm{H}_{3} \mathrm{Cl}\). To find the molecular weight, we'll add the atomic weights of the constituent elements: - Carbon (C) has an atomic weight of 12.01 g/mol - Hydrogen (H) has an atomic weight of 1.008 g/mol - Chlorine (Cl) has an atomic weight of 35.45 g/mol So, the molecular weight of vinyl chloride is: (2 x 12.01) + (3 x 1.008) + 35.45 = 24.02 + 3.024 + 35.45 = 62.494 g/mol
02

Calculate the moles of vinyl chloride per liter

We know that the maximum allowable concentration of vinyl chloride is \(2.0 \times 10^{-6} \ \mathrm{g}/\mathrm{L}\). To get the number of moles per liter, we will divide the mass of vinyl chloride by its molecular weight: \[\text{Moles per liter} = \frac{2.0 \times 10^{-6}\ \mathrm{g}}{62.494\ \mathrm{g/mol}}\] Calculate the division: \[\text{Moles per liter} \approx 3.20 \times 10^{-8}\ \mathrm{mol/L}\]
03

Calculate the number of molecules per liter

To find the number of molecules per liter, we'll use Avogadro's number, which is approximately \(6.022 \times 10^{23}\ \mathrm{molecules/mol}\). Multiply this by the number of moles per liter: \[\text{Molecules per liter} = (3.20 \times 10^{-8}\ \mathrm{mol/L}) \times (6.022 \times 10^{23}\ \mathrm{molecules/mol})\] Calculate the multiplication: \[\text{Molecules per liter} \approx 1.93 \times 10^{16}\ \mathrm{molecules/L}\]
04

Final answers

To summarize, the allowable concentration level of vinyl chloride \(\mathrm{C}_{2}\mathrm{H}_{3} \mathrm{Cl}\) in the atmosphere represents: 1. Approximately \(3.20 \times 10^{-8}\) moles per liter 2. Approximately \(1.93 \times 10^{16}\) molecules per liter.

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.

Molecular Weight Calculation
The molecular weight of a compound is the sum of the atomic weights of all the atoms in its molecular formula. In our case, we are looking at vinyl chloride, which has the molecular formula \(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{Cl}\). To find the molecular weight, simply add up the atomic weights of each element in the formula, multiplied by the number of atoms of that element present:
  • Carbon (\(\mathrm{C}\)) has an atomic weight of 12.01 g/mol and there are 2 atoms of it, so \(2 \times 12.01 = 24.02 \) g/mol.
  • Hydrogen (\(\mathrm{H}\)) has an atomic weight of 1.008 g/mol and there are 3 atoms, so \(3 \times 1.008 = 3.024 \) g/mol.
  • Chlorine (\(\mathrm{Cl}\)) has an atomic weight of 35.45 g/mol and there is 1 atom, so \(1 \times 35.45 = 35.45 \) g/mol.
By adding these values together, we get the molecular weight: \(24.02 + 3.024 + 35.45 = 62.494 \) g/mol. Understanding how to calculate molecular weight is crucial in chemistry as it helps in converting between grams and moles, which are essential for quantitative chemical analysis.
Avogadro's Number
Avogadro's number is a fundamental constant that helps us understand the scale at which chemical processes operate. It is approximately \(6.022 \times 10^{23}\), and it represents the number of atoms, ions, or molecules contained in one mole of a substance. This number is named after the Italian scientist Amedeo Avogadro.Why is Avogadro's number important? It allows chemists to "count" very large numbers of tiny particles by relating the macroscopic measurements we can make in the lab to the microscopic count of individual molecules or atoms:
  • When you know the number of moles of a substance, you can multiply it by Avogadro's number to find the number of particles.
  • Conversely, if you know how many molecules you have, you can divide by Avogadro's number to get the number of moles.
This immense figure portrays the size of even tiny quantities of substances, providing an essential bridge from the tangible world to the molecular scale, allowing calculations that make practical chemistry possible.
Mole-to-Molecule Conversion
Converting from moles to molecules (or vice versa) is an essential skill in chemistry, as it helps you quantify the number of particles in a given amount of substance. To make this conversion, Avogadro's number is the key factor.For example, if you think of moles as the "scientific dozen," then just like a dozen represents 12 objects, a mole represents \(6.022 \times 10^{23}\) molecules. Here's how you can use it:
  • **Convert from moles to molecules**: Multiply the number of moles by Avogadro's number. For instance, with vinyl chloride, if you have \(3.20 \times 10^{-8}\) moles, you calculate the number of molecules by multiplying \((3.20 \times 10^{-8} \times 6.022 \times 10^{23})\) to get approximately \(1.93 \times 10^{16}\) molecules.
  • **Convert from molecules to moles**: Divide the number of molecules by Avogadro's number.
This step is crucial when you need to understand the proportion of molecules involved in reactions or the concentration levels in solutions. Mastery of this conversion allows for deeper comprehension of chemical reactions on a molecular level.

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

Balance the following equations: (a) \(\mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)\) (b) \(\mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HNO}_{3}(a q)\) (c) \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(l)+\mathrm{HCl}(g)\) (d) \(\mathrm{Al}_{4} \mathrm{C}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{CH}_{4}(g)\) (e) \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (f) \(\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (g) \(\mathrm{Mg}_{3} \mathrm{~N}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) \(\mathrm{MgSO}_{4}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(a q)\)

A sample of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) contains \(1.250 \times 10^{21}\) carbon atoms. (a) How many atoms of hydrogen does it contain? (b) How many molecules of glucose does it contain? (c) How many moles of glucose does it contain? (d) What is the mass of this sample in grams?

Write balanced chemical equations to correspond to each of the following descriptions: (a) Solid calcium carbide, \(\mathrm{CaC}_{2}\), reacts with water to form an aqueous solution of calcium hydroxide and acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}\). (b) When solid potassium chlorate is heated, it decomposes to form solid potassium chloride and oxygen gas. (c) Solid zinc metal reacts with sulfuric acid to form hydrogen gas and an aqueous solution of zinc sulfate. (d) When liquid phosphorus trichloride is added to water, it reacts to form aqueous phosphorous acid, \(\mathrm{H}_{3} \mathrm{PO}_{3}(a q),\) and aqueous hydrochloric acid. (e) When hydrogen sulfide gas is passed over solid hot iron(III) hydroxide, the resultant reaction produces solid iron(III) sulfide and gaseous water.

Aspirin \(\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\right)\) is produced from salicylic acid \(\left(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}\right)\) and acetic anhydride \(\left(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}\right):\) $$ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}+\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3} \longrightarrow \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} $$ (a) How much salicylic acid is required to produce \(1.5 \times 10^{2} \mathrm{~kg}\) of aspirin, assuming that all of the salicylic acid is converted to aspirin? (b) How much salicylic acid would be required if only \(80 \%\) of the salicylic acid is converted to aspirin? (c) What is the theoretical yield of aspirin if \(185 \mathrm{~kg}\) of salicylic acid is allowed to react with \(125 \mathrm{~kg}\) of acetic anhydride? (d) If the situation described in part (c) produces \(182 \mathrm{~kg}\) of aspirin, what is the percentage yield?

Write a balanced chemical equation for the reaction that occurs when (a) \(\mathrm{Mg}(s)\) reacts with \(\mathrm{Cl}_{2}(g) ;\) (b) barium carbonate decomposes into barium oxide and carbon dioxide gas when heated; \((\mathbf{c})\) the hydrocarbon styrene, \(\mathrm{C}_{8} \mathrm{H}_{8}(l)\), is combusted in air; (d) dimethylether, \(\mathrm{CH}_{3} \mathrm{OCH}_{3}(g)\), is combusted in air.

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