Let's go through the problem and solution step-by-step:
1. Understanding the Problem:
We are given:
- Mass ([tex]\( m \)[/tex]) = 3.1 kg
- Density ([tex]\( \rho \)[/tex]) = 514 g/L
We need to calculate the volume ([tex]\( V \)[/tex]) that this mass would occupy, given the density.
2. Convert Units:
The density is given in grams per liter (g/L), but the mass is in kilograms (kg). To have consistent units, we convert the mass from kilograms to grams.
Conversion factor: [tex]\( 1 \text{ kg} = 1000 \text{ g} \)[/tex]
[tex]\[
m = 3.1 \text{ kg} \times 1000 \frac{\text{g}}{\text{kg}} = 3100 \text{ g}
\][/tex]
3. Formula for Volume:
The formula for density is:
[tex]\[
\rho = \frac{m}{V}
\][/tex]
Rearrange this formula to solve for volume:
[tex]\[
V = \frac{m}{\rho}
\][/tex]
4. Substitute the Values:
Substitute the given mass (in grams) and the density into the rearranged formula:
[tex]\[
V = \frac{3100 \text{ g}}{514 \frac{\text{g}}{\text{L}}}
\][/tex]
5. Calculate the Volume:
Using a calculator, perform the division:
[tex]\[
V = 6.031128404669261 \text{ L}
\][/tex]
6. Rounding to Significant Digits:
According to the significant figures rule, the volume should be reported to match the precision of the given data. Both the mass (3.1) and density (514) have 3 significant figures. We round our result to 3 significant figures:
[tex]\[
V \approx 6.031 \text{ L}
\][/tex]
Therefore, the volume of 3.1 kg of the liquid with a density of 514 g/L is approximately [tex]\(6.031 \text{ L}\)[/tex].
