To solve this question, we need to utilize the relationship between mass, density, and volume. The formula we use here is:
[tex]\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \][/tex]
Given:
- The density of bromine ([tex]\( Br_2 \)[/tex]) is [tex]\( 3.12 \text{ g/cm}^3 \)[/tex].
- The volume of bromine in the first part is [tex]\( 125 \text{ mL} \)[/tex]. Note that 1 mL is equivalent to 1 cm³.
- The mass of bromine in the second part is [tex]\( 85.0 \text{ g} \)[/tex].
### 1. Calculate the mass of 125 mL of bromine
We are given the volume and the density, and we need to find the mass. Rearrange the formula for mass:
[tex]\[ \text{Mass} = \text{Density} \times \text{Volume} \][/tex]
Now, substitute the known values:
[tex]\[ \text{Mass} = 3.12 \text{ g/cm}^3 \times 125 \text{ cm}^3 \][/tex]
Calculate the mass:
[tex]\[ \text{Mass} = 3.12 \times 125 = 390.0 \text{ g} \][/tex]
Therefore, the mass of 125 mL of bromine is [tex]\( 390.0 \text{ g} \)[/tex].
### 2. Calculate the volume that 85.0 g of bromine occupies
Given the mass and the density, we need to find the volume. Rearrange the formula for volume:
[tex]\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \][/tex]
Now, substitute the known values:
[tex]\[ \text{Volume} = \frac{85.0 \text{ g}}{3.12 \text{ g/cm}^3} \][/tex]
Calculate the volume:
[tex]\[ \text{Volume} = \frac{85.0}{3.12} \approx 27.24358974358974 \text{ cm}^3 \][/tex]
Therefore, the volume occupied by 85.0 g of bromine is approximately [tex]\( 27.24 \text{ cm}^3 \)[/tex] (rounded to two decimal places).
### Summary of Results:
- The mass of 125 mL of bromine is [tex]\( 390.0 \text{ g} \)[/tex].
- The volume that 85.0 g of bromine occupies is approximately [tex]\( 27.24 \text{ cm}^3 \)[/tex].
