Seawater contains 0.0065% (by mass) of bromine. How many grams of bromine are there in 2.50 L of seawater? The density of seawater is 1.025 g/cm^3
Solution:
Let us first find the mass of the seawater.
m_{seawater}=2.50L\times \frac{10^3\text{cm}^3}{1L}\times \frac{1.025g}{1\text{cm}^3}=2.5625\times 10^3\,g
Now, we can find the mass percentage of bromine.
Mass percentage = \frac{\text{mass of Br in seawater}}{\text{mass of seawater}}\times 100\%
Isolating for mass of bromine gives us:
Mass Br =\frac{0.0065\%}{100\%}\times 2.5625\times 10^3\,g
Mass Br=0.166 g
This question can be found in General Chemistry, 9th edition, chapter 3, question 3.54
