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 0.998g/mL? (1 ppb Hg = 1ng Hg/g water)

0.68ppb:______g

a)0.015g

b)0.010g

c)0.020g

d)0.025g

Answer 1

To calculate the mercury content in 15.0 L of water with a concentration of 0.68 ppb, convert the volume to mass, then multiply by the mercury concentration, and convert from nanograms to grams. The result is 0.010 g of mercury, which is option (b).

Step-by-step explanation:

To calculate the quantity of mercury in grams present in 15.0 L of water with a mercury concentration of 0.68 parts per billion (ppb), follow these steps:

First, convert the volume of water from liters to milliliters as the density is given in g/mL. So, 15.0 L × 1000 mL/L = 15000 mL.

Now, convert the water's volume to mass using the given density (0.998 g/mL). Mass = volume × density, which gives us 15000 mL × 0.998 g/mL = 14970 g.

Since 1 ppb corresponds to 1 ng/g, a concentration of 0.68 ppb is equivalent to 0.68 ng/g. To find the total mass of mercury, multiply the mass of the water by the concentration of mercury: 14970 g × 0.68 ng/g = 10179.6 ng.

Finally, convert the mass of mercury from nanograms to grams by dividing by 10^9 (since there are 10^9 ng in a g). This gives us 10179.6 ng / 10^9 ng/g = 0.0101796 g, which can be approximated to 0.010g when rounded to three significant digits.

Hence, the quantity of mercury in grams present in 15.0 L of the water is 0.010 g, which corresponds to option (b) 0.010g.