Certainly! Let's find the volume that 35.2 grams of carbon tetrachloride ([tex]$CCl_4$[/tex]) will occupy, given its density of [tex]$1.60 \, \text{g/mL}$[/tex].
We can use the formula for density to find the volume. The formula for density (ρ) is given by:
[tex]\[
\rho = \frac{m}{V}
\][/tex]
where:
- [tex]\( \rho \)[/tex] is the density,
- [tex]\( m \)[/tex] is the mass, and
- [tex]\( V \)[/tex] is the volume.
We can rearrange this formula to solve for volume (V):
[tex]\[
V = \frac{m}{\rho}
\][/tex]
Given:
- [tex]\( m = 35.2 \, \text{g} \)[/tex]
- [tex]\( \rho = 1.60 \, \text{g/mL} \)[/tex]
Substitute the given values into the formula:
[tex]\[
V = \frac{35.2 \, \text{g}}{1.60 \, \text{g/mL}}
\][/tex]
When we perform this division, we get:
[tex]\[
V = 22.0 \, \text{mL}
\][/tex]
Therefore, the volume that 35.2 grams of carbon tetrachloride ([tex]$CCl_4$[/tex]) will occupy is:
[tex]\[
22.0 \, \text{mL}
\][/tex]
