Question

The mercury content of a stream was believed to be above the minimum considered safe—1 part per billion (ppb) by weight. An analysis indicated that the concentration was 0.68 parts per billion. What quantity of mercury in grams was present in 15.0 L of the water, the density of which is 0.998 g/ml? (1 ppb Hg=1 ng Hg1 g water)

Answer 1

1.02 x 10⁻⁵ g mercury

Step-by-step explanation:

Given the concentration in ppb ( 1 ng = 1g H₂O ), we need to first calculate the mass of water of 15 L of water. We are given the density so:

d = m/V ⇒ m = d x V

We need to convert the volume to mL to be consistent in the units:

V = 15.0 L x 1000 mL / 1L = 15000 mL

Now, the mass of water is

m = 0.998 g/mL x 15000 mL = 1.497 x 10⁴ g H₂O

Now we can calculate the quantity of mercury from the given concentration in ppb:

1.497 x 10⁴ g H₂O x 0.68 ng mercury / g H₂O = 1.018 x 10⁴ ng mercury

We need to convert this mass to g since we are asked to calculate the quantity of mercury in g.

mass mercury = 1.018 x 10⁴ ng x ( 1 g / 1 x 10⁹ ng ) ( 1 g = 10⁹ ng )

= 1.02 x 10⁻⁵ g ( to three significant figures )