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Chapter 3: Problem 84

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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Most popular questions from this chapter

Chlorophyll, the substance responsible for the green color of leaves, has one magnesium atom per chlorophyll molecule and contains \(2.72 \%\) magnesium by mass. What is the molar mass of chlorophyll?

97\. Answer the questions below, using LT (for is less than), GT (for is greater than), EQ (for is equal to), or MI (for more information required) in the blanks provided. (a) The mass (to three significant figures) of \(6.022 \times 10^{23}\) atoms of Na ______ 23.0 g. (b) Boron has two isotopes, B-10 (10.01 amu) and B-11 (11.01 amu). The abundance of B-10 _______the abundance of B-11. (c) If S-32 were assigned as the standard for expressing relative atomic masses and assigned an atomic mass of \(10.00 \mathrm{amu}\), the atomic mass for \(\mathrm{H}\) would be _____\(1.00 \mathrm{amu} .\) (d) When phosphine gas, \(\mathrm{PH}_{3}\), is burned in oxygen, tetraphosphorus decaoxide and steam are formed. In the balanced equation (using smallest whole-number coefficients) for the reaction, the sum of the coefficients on the reactant side is _______ \(7 .\) (e) The mass (in grams) of one mole of bromine molecules is _______ \(79.90\)

Convert the following to moles. (a) \(35.00 \mathrm{~g}\) of \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\), a chlorofluorocarbon that destroys the ozone layer in the atmosphere (b) \(100.0 \mathrm{mg}\) of iron(II) sulfate, an iron supplement prescribed for anemia (c) \(2.00 \mathrm{~g}\) of Valium \(\left(\mathrm{C}_{15} \mathrm{H}_{13} \mathrm{ClN}_{2} \mathrm{O}-\right.\) diazepam \()\)

A sample of an oxide of vanadium weighing \(4.589 \mathrm{~g}\) was heated with hydrogen gas to form water and another oxide of vanadium weighing \(3.782 \mathrm{~g} .\) The second oxide was treated further with hydrogen until only \(2.573 \mathrm{~g}\) of vanadium metal remained. (a) What are the simplest formulas of the two oxides? (b) What is the total mass of water formed in the successive reactions?

Consider the hypothetical reaction $$ 8 \mathrm{~A}_{2} \mathrm{~B}_{3}(s)+3 \mathrm{X}_{4}(g) \longrightarrow 4 \mathrm{~A}_{4} \mathrm{X}_{3}(s)+12 \mathrm{~B}_{2}(g) $$ When \(10.0 \mathrm{~g}\) of \(\mathrm{A}_{2} \mathrm{~B}_{3}(\mathrm{MM}=255 \mathrm{~g} / \mathrm{mol})\) react with an excess of \(\mathrm{X}_{4}, 4.00 \mathrm{~g}\) of \(\mathrm{A}_{4} \mathrm{X}_{3}\) are produced. (a) How many moles of \(A_{4} X_{3}\) are produced? (b) What is the molar mass of \(\mathrm{A}_{4} \mathrm{X}_{3}\) ?
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