Answer :
Sure! To find out how many molecules are contained in 98.3 g of [tex]\( \text{CCl}_4 \)[/tex] (carbon tetrachloride), follow these steps:
1. Calculate the Molar Mass of [tex]\( \text{CCl}_4 \)[/tex]:
- Carbon (C) has an atomic mass of approximately 12.01 g/mol.
- Chlorine (Cl) has an atomic mass of approximately 35.45 g/mol.
- Since [tex]\( \text{CCl}_4 \)[/tex] consists of one carbon atom and four chlorine atoms, the molar mass is:
[tex]\[
12.01 + 4 \times 35.45 = 153.81 \, \text{g/mol}
\][/tex]
2. Determine the Number of Moles of [tex]\( \text{CCl}_4 \)[/tex]:
- Use the formula for moles: [tex]\(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\)[/tex].
- Substitute the given mass and the molar mass:
[tex]\[
\text{moles of } \text{CCl}_4 = \frac{98.3 \, \text{g}}{153.81 \, \text{g/mol}} \approx 0.6391 \, \text{mol}
\][/tex]
3. Calculate the Number of Molecules:
- Use Avogadro’s number, which is approximately [tex]\( 6.022 \times 10^{23} \)[/tex] molecules/mol, to find the number of molecules.
- Multiply the number of moles by Avogadro's number:
[tex]\[
\text{molecules of } \text{CCl}_4 = 0.6391 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \approx 3.85 \times 10^{23} \, \text{molecules}
\][/tex]
Therefore, the number of molecules contained in 98.3 g of [tex]\( \text{CCl}_4 \)[/tex] is approximately [tex]\( 3.85 \times 10^{23} \)[/tex] molecules. The correct answer is [tex]\( 3.85 \times 10^{23} \)[/tex] molecules.
1. Calculate the Molar Mass of [tex]\( \text{CCl}_4 \)[/tex]:
- Carbon (C) has an atomic mass of approximately 12.01 g/mol.
- Chlorine (Cl) has an atomic mass of approximately 35.45 g/mol.
- Since [tex]\( \text{CCl}_4 \)[/tex] consists of one carbon atom and four chlorine atoms, the molar mass is:
[tex]\[
12.01 + 4 \times 35.45 = 153.81 \, \text{g/mol}
\][/tex]
2. Determine the Number of Moles of [tex]\( \text{CCl}_4 \)[/tex]:
- Use the formula for moles: [tex]\(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\)[/tex].
- Substitute the given mass and the molar mass:
[tex]\[
\text{moles of } \text{CCl}_4 = \frac{98.3 \, \text{g}}{153.81 \, \text{g/mol}} \approx 0.6391 \, \text{mol}
\][/tex]
3. Calculate the Number of Molecules:
- Use Avogadro’s number, which is approximately [tex]\( 6.022 \times 10^{23} \)[/tex] molecules/mol, to find the number of molecules.
- Multiply the number of moles by Avogadro's number:
[tex]\[
\text{molecules of } \text{CCl}_4 = 0.6391 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \approx 3.85 \times 10^{23} \, \text{molecules}
\][/tex]
Therefore, the number of molecules contained in 98.3 g of [tex]\( \text{CCl}_4 \)[/tex] is approximately [tex]\( 3.85 \times 10^{23} \)[/tex] molecules. The correct answer is [tex]\( 3.85 \times 10^{23} \)[/tex] molecules.
