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Chapter 3: Problem 43

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 \(2.0 \times 10^{-6} \mathrm{g} / \mathrm{L}\) of vinyl chloride is equivalent to \(3.2 \times 10^{-8}\) moles per liter and \(1.928 \times 10^{16}\) molecules of vinyl chloride per liter.

Step by step solution

01

Calculate the molar mass of vinyl chloride

First, we need to find the molar mass of vinyl chloride (C2H3Cl). We can do this by summing the atomic masses of the elements in the molecule using the periodic table. The molar mass of carbon (C) is 12.01 g/mol, the molar mass of hydrogen (H) is 1.01 g/mol, and the molar mass of chlorine (Cl) is 35.45 g/mol. C2H3Cl = 2(12.01) + 3(1.01) + 35.45 = 24.02 + 3.03 + 35.45 = 62.50 g/mol
02

Calculate the moles of vinyl chloride in each liter

Now that we have the molar mass, we can calculate the moles of vinyl chloride in each liter. We can use the formula: moles = (mass of substance)/(molar mass) The given mass of vinyl chloride is \(2.0 \times 10^{-6}\) g/L. Using the molar mass calculated in Step 1, we now have: moles = (\(2.0 \times 10^{-6}\) g/L) / (62.50 g/mol) moles = \(3.2 \times 10^{-8}\) mol/L
03

Calculate the number of molecules per liter

To find the number of molecules per liter, we can multiply the moles per liter by Avogadro's number which is approximately \(6.022 \times 10^{23}\) molecules/mol. number of molecules = (moles per liter) × (Avogadro's number) number of molecules = (\(3.2 \times 10^{-8}\) mol/L) × (\(6.022 \times 10^{23}\) molecules/mol) number of molecules = \(1.928 \times 10^{16}\) molecules/L
04

Present the final results

From our calculations, we find that the allowable concentration level of vinyl chloride in the atmosphere in a chemical plant, \(2.0 \times 10^{-6} \mathrm{g} / \mathrm{L}\), is equivalent to: 1. \(3.2 \times 10^{-8}\) moles of vinyl chloride per liter 2. \(1.928 \times 10^{16}\) molecules of vinyl chloride per liter

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Most popular questions from this chapter

The combustion of one mole of liquid ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) , produces 1367 \(\mathrm{kJ}\) of heat. Calculate how much heat is produced when 235.0 \(\mathrm{g}\) of ethanol are combusted.

A key step in balancing chemical equations is correctly identifying the formulas of the reactants and products. For example, consider the reaction between calcium oxide, CaO(s), and \(\mathrm{H}_{2} \mathrm{O}(l)\) to form aqueous calcium hydroxide. (a) Write a balanced chemical equation for this combination reaction, having correctly identified the product as \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)\) . (b) Is it possible to balance the equation if you incorrectly identify the product as \(\mathrm{CaOH}(a q),\) and if so, what is the equation?

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Balance the following equations: $$ \begin{array}{l}{\text { (a) } \mathrm{Al}_{4} \mathrm{C}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{CH}_{4}(g)} \\ {\text { (b) } \mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (c) } \mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)} \\ {\text { (d) } \mathrm{Mg}_{3} \mathrm{N}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{MgSO}_{4}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(a q)}\end{array} $$
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