Question

The mercury content of a stream was belived 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 persent in 15.0 L of the water, the density of which is 0.998 g/mL? (1 ppb Hg = 1ngHg/1g water)

Answer 1

Quantity of Hg in 15.00 L water = 1.0*10^-5 g

Step-by-step explanation:

Given:

Concentration of Hg = 0.68 parts per billion (ppb)

Volume of water = 15.0 L

Density of water = 0.998 g/ml

To determine:

The amount of Hg in 15.0 L of water

Calculation:

1 ppb = 1 nanogram (ng) solute/1 g water

1 ng = 10⁻⁹ g

0.68 ppb = 0.68 ngHg/1 g water = 0.68*10⁻⁹g Hg/ 1g water

`Mass\ of\ water = Density*Volume \\\\= 0.998 g/ml*15.0*10^(3) ml = 15.0*10^(3) g`

The amount of Hg in the above mass of water would be:

=
`(15.0*10^(3)g\ water*0.68*10^(-9)g\ Hg  )/(1g\ water) =1.0*10^(-5) g`