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Chapter 1: Q86E (page 9)

Question: The mercury content of a stream was believed to be above the minimum considered safe 鈥� 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? \(\left( {1ppbHg = \frac{{1ngHg}}{{1gwater}}} \right)\)

Short Answer

Expert verified

10.18 x 10-6 g

Step by step solution

01

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

02

Compute the weight of Hg

1ppbofHg=1gHg1gwater=1脳10-9gHg1gwater0.68=x脳10914970g=x脳10614.97x=0.68脳14.97106=10.18脳10-6g

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