Answer:
[tex]\boxed{\text{(3) 9.6 L}}[/tex]
Explanation:
1. Moles of CCl₄
[tex]n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}[/tex]
2. Molar mass of CCl₄
MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol
3. Mass of CCl₄
[tex]m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}[/tex]
4. Volume of CCl₄
[tex]V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}[/tex]
