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Carbon tetrachloride, \(\mathrm{CCl}_{4}\), was a popular dry-cleaning agent until it was shown to be carcinogenic. It has a density of \(1.589 \mathrm{~g} / \mathrm{cm}^{3} .\) What volume of carbon tetrachloride will contain a total of \(6.00 \times 10^{25}\) molecules of \(\mathrm{CCl}_{4} ?\)

Short Answer

Expert verified
Answer: The volume of carbon tetrachloride containing \(6.00 \times 10^{25}\) molecules of \(\mathrm{CCl}_{4}\) is \(9,646.90 \mathrm{cm^{3}}\).

Step by step solution

01

Find the molar mass of \(\mathrm{CCl}_{4}\)

First, determine the molar mass of carbon tetrachloride. Carbon has a molar mass of \(12.01 \mathrm{~g/mol}\) and Chlorine has a molar mass of \(35.45 \mathrm{~g/mol}\). So, the molar mass of \(\mathrm{CCl}_{4}\) is: \(1(12.01 \mathrm{~g/mol}) + 4(35.45 \mathrm{~g/mol}) = 12.01 + 4(35.45) = 12.01 + 141.8 = 153.81 \mathrm{~g/mol}\)
02

Determine the number of moles of \(\mathrm{CCl}_{4}\)

Next, we need to find the number of moles of \(\mathrm{CCl}_{4}\). We can use Avogadro's number (\(6.022 \times 10^{23} \mathrm{molecules/mol})\) to find the number of moles: Number of moles = (Number of molecules given) / (Avogadro's number) Number of moles = (\(6.00 \times 10^{25} \mathrm{molecules})\) / (\(6.022 \times 10^{23} \mathrm{molecules/mol}) = 99.63 \mathrm{mol}\)
03

Calculate the mass of \(\mathrm{CCl}_{4}\)

Now we will find the mass of carbon tetrachloride containing the given molecules. This can be found using the formula: Mass = (number of moles) × (molar mass) Mass = (99.63 \mathrm{mol}) × (153.81 \mathrm{~g/mol}) = 15,335.37 \mathrm{g}$
04

Calculate the volume of \(\mathrm{CCl}_{4}\)

Finally, we will use the density formula to find the volume of carbon tetrachloride containing the given molecules: Volume = (mass) / (density) Volume = \((15,335.37 \mathrm{g})\) / \((1.589 \mathrm{g/cm^{3}})\) = 9,646.90 \mathrm{cm^{3}}$ The volume of carbon tetrachloride containing \(6.00 \times 10^{25}\) molecules of \(\mathrm{CCl}_{4}\) is \(9,646.90 \mathrm{cm^{3}}\).

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Key Concepts

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

Molar Mass Calculation
In chemical calculations, understanding molar mass is crucial. It allows us to connect the molecular world with the mass we can measure. Molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol). To find the molar mass of a compound like carbon tetrachloride (\mathrm{CCl}_4):
  • Identify the molar mass of each element in the compound.
  • Carbon (C) has a molar mass of 12.01 g/mol.
  • Chlorine (Cl) has a molar mass of 35.45 g/mol.
Combine these values according to the number of each atom in the compound. For \mathrm{CCl}_4, there is 1 carbon atom and 4 chlorine atoms:\[Molar\ mass\ of\ CCl_4\ = 1(12.01\ g/mol) + 4(35.45\ g/mol) = 153.81\ g/mol\]Knowing the molar mass allows us to convert between grams and moles, making it a powerful tool in chemistry.
Avogadro's Number
Avogadro’s number is a fundamental constant in chemistry used to connect the number of particles, like atoms or molecules, to the amount of substance in moles. This conversion is essential when dealing with large numbers of tiny particles. Avogadro's number is:
  • 6.022 \times 10^{23} molecules/mol.
To find the number of moles from the number of molecules, use the formula:\[\text{Number of moles} = \frac{\text{Number of molecules}}{\text{Avogadro's number}}\]For example, with 6.00 \times 10^{25} molecules of \mathrm{CCl}_4, calculating moles involves:\[\text{Number of moles} = \frac{6.00 \times 10^{25}\ \text{molecules}}{6.022 \times 10^{23}\ \text{molecules/mol}} = 99.63\ \text{mol}\]This calculation is the key to understanding how many moles of a substance you have based on its molecules.
Density and Volume Relationship
Density is a measure of how much mass is contained in a given volume. For chemical calculations, density helps convert mass to volume or vice versa. The formula that connects them is:\[Volume = \frac{\text{Mass}}{\text{Density}}\]In this problem, we use the density of carbon tetrachloride, 1.589\ g/cm^3, to find its volume given a mass. Here’s how it works:
  • Calculate the mass of the molecules using mass = moles \times molar mass.
  • Determine the volume using the density formula above.
From the given 15,335.37\ g mass of \mathrm{CCl}_4, the volume is:\[Volume = \frac{15,335.37\ g}{1.589\ g/cm^3} = 9,646.90\ cm^3\]Thus, understanding density and its relation with volume provides a way to move between these physical quantities.

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Most popular questions from this chapter

Consider an isotope of yttrium, Y-90. This isotope is incorporated into cancer-seeking antibodies so that the cancer can be irradiated by the yttrium and destroyed. How many neutrons are in (a) twenty-five atoms of yttrium? (b) \(0.250\) mol of yttrium? (c) one nanogram \(\left(10^{-9} \mathrm{~g}\right)\) of yttrium?

When potassium chlorate is burned, potassium chloride and oxygen are formed. (a) Write a balanced equation for the reaction. (b) How much potassium chlorate must be burned to produce \(198.5 \mathrm{~g}\) of oxygen? The yield is found to be \(83.2 \%\).

Determine the simplest formulas of the following compounds: (a) saccharin, the artificial sweetener, which has the composition \(45.90 \% \mathrm{C}, 2.75 \% \mathrm{H}, 26.20 \% \mathrm{O}, 17.50 \% \mathrm{~S}\), and \(7.65 \% \mathrm{~N}\) (b) allicin, the compound that gives garlic its characteristic odor, which has the composition \(6.21 \% \mathrm{H}, 44.4 \% \mathrm{C}, 9.86 \% \mathrm{O}\), and \(39.51 \% \mathrm{~S}\). (c) sodium thiosulfate, the fixer used in developing photographic film, which has the composition \(30.36 \% \mathrm{O}, 29.08 \% \mathrm{Na}\), and \(40.56 \% \mathrm{~S}\).

Balance the following equations: (a) \(\mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{S}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(\mathrm{CH}_{4}(g)+\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{HCN}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{Fe}(l)+\mathrm{H}_{2} \mathrm{O}(g)\)

Calculate the mass in grams of \(2.688 \mathrm{~mol}\) of (a) chlorophyll, \(\mathrm{C}_{55} \mathrm{H}_{72} \mathrm{~N}_{4} \mathrm{O}_{5} \mathrm{Mg}\), responsible for the green color of leaves. (b) sorbitol, \(\mathrm{C}_{9} \mathrm{H}_{14} \mathrm{O}_{6}\), an artificial sweetener. (c) indigo, \(\mathrm{C}_{16} \mathrm{H}_{10} \mathrm{~N}_{2} \mathrm{O}_{2}\), a blue dye.

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