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 hg ) 1 g water

Answer 1

Answer:

Amount of mercury is 1.0*10⁻⁵ g

Explanation:

Given:

Mercury content of stream = 0.68 ppb

volume of water = 15.0 L

Density of water = 0.998 g/L

To determine:

Amount of mercury in 15.0 L of water

Calculation:

$1 ppb = \frac{1\mu g(solute)}{1L(solvent)}$

where 1 μg (micro gram) = 10⁻⁶ g

0.68 ppm implies that there is 0.68 *10⁻⁶ g mercury per Liter of water

Therefore, the amount of mercury in 15.0 L water would be:

$=\frac{0.68*10^{-6}g\ Hg* 15.0\ L\ water}{1\ L\ water} =1.02*10^{-5}g$