Answer
There is 0.17 g bromine in 2.5 L seawater.
Work Step by Step
There is 0.0065% (by mass) bromine in seawater. This means that in the given mass of the seawater sample 0.0065% should be bromine.
We are given the volume of the sample and the density of seawater, so we can find the mass of the seawater sample:
m$_{seawater sample}$ = V x d
The volume of seawater sample is 2.5 L or 2.5 x $10^{3}$ mL
The density of seawater is 1.025 g/$cm^{3}$ or 1.025 g/mL ( 1$cm^{3}$ = 1 mL) .
m$_{seawater sample}$ = V x d
m$_{seawater sample}$ =2.5 x $10^{3}$ mL x 1.025 g/mL
m$_{seawater sample}$ = 2562.5 g
Now we find the mass of bromine in this sample of seawater:
bromine mass = 0.0065% x mass of the seawater sample
bromine mass = 0.0065% x 2562.5 g
bromine mass = $\frac{0.0065}{100}$ x 2562.5 g
bromine mass = 0.1665 g
bromine mass= 0.17 g
There is 0.17 g bromine in 2.5 L seawater.
