Question

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

The volume of carbon tetrachloride that contains 6.00 x 10^25 molecules is 9.63 x 10^3 cm^3.

Step-by-step solution

Calculate the number of moles

To convert molecules to moles, we need to divide the given number of molecules by Avogadro's number, which is \(6.022 \times 10^{23}\) molecules per mole: $$ \text{Number of moles} = \frac{6.00 \times 10^{25}\text{ molecules}}{6.022 \times 10^{23}\text{ molecules/mol}} = 9.96 \times 10^{1}\text{ mol} $$

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

To calculate the mass, we need to multiply the number of moles by the molar mass of \(\mathrm{CCl}_{4}\). The molar mass of \(\mathrm{CCl}_{4}\) is: $$ \text{Molar mass} =1 \times 12.01~(C) \mathrm{g/mol} +4 \times 35.45~(\text{Cl}) \mathrm{g/mol} = 153.8\mathrm{~g/mol} $$ So, the mass of \(\mathrm{CCl}_{4}\) is: $$ \text{Mass} = 9.96 \times 10^{1}\text{ mol} \times 153.8 \mathrm{~g/mol} = 1.53 \times 10^{4} \mathrm{~g} $$

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

To calculate the volume of carbon tetrachloride, we will use its density, which is given as \(1.589\mathrm{~g}/\mathrm{cm}^{3}\). Using the mass we calculated in Step 2 and the density, we can find the volume: $$ \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{1.53 \times 10^{4} \mathrm{~g}}{1.589 \mathrm{~g}/\mathrm{cm}^{3}} = 9.63 \times 10^{3}\text{ cm}^{3} $$ So, the volume of carbon tetrachloride that contains \(6.00 \times 10^{25}\) molecules is \(9.63 \times 10^{3}\text{ cm}^{3}\).