Q86E - Chemistry (OpenStax)

Question

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? \(\left( {1ppbHg = \frac{{1ngHg}}{{1gwater}}} \right)\)

Step-by-Step Solution

Answer

10.18 x 10^-6 g

Step 1: Determine the weight of the stream

\(Density = \frac{m}{V}\)

Where m= mass

V = Volume

The density of the stream is given = 0.998 g/ml

This implies that 1 ml stream weighs 0.998 g.

Now, the volume of the water given = 15 L= 15000 mL

So, the weight of 15000 mL stream =  0.998/1ml*15000 mL = 14970 g

Step 2: Compute the weight of Hg

$$\begin{gathered} 1 ppb of Hg = \frac{1g Hg}{1g water}=\frac{1\times {10}^{-9}g Hg}{1 g water} \\ 0.68=\frac{x\times {10}^9}{14970 g}=\frac{x\times {10}^6}{14.97} \\ x=\frac{0.68\times 14.97}{{10}^6}=10.18\times {10}^{-6} g \end{gathered}$$