Question

Using the highest value of lead from the first draw data set, and assuming a resident drinks 3.5 L of water per day, calculate the mass of lead that the resident would consume over the course of one year. (Assume the water has a density of 1.0 g/mL) Express your answer to two significant figures.

First, calculate the mass of water consumed in one year. Then, use the highest lead level in the first-draw method's data set to calculate the mass of lead found in that mass of water. The assumed density of water, 1.0 g/mL , can also be written as 1000 g/L . The unit ppb is equivalent to (g solute)/(109 g solvent)(g solute)/(109 g solvent).

Sample number	Lead level first draw (ppbppb)	Lead level 45-ssflush (ppbppb)	Lead level 2-minminflush (ppbppb)
1	0.344	0.226	0.145
2	8.133	10.77	2.761
3	1.111	0.11	0.123
4	8.007	7.446	3.384
5	1.951	0.048	0.035
6	7.2	1.4	0.2
7	40.63	9.726	6.132
8	1.1	2.5	0.1
9	10.6	1.038	1.294
10	6.2	4.2	2.3
11	4.358	0.822	0.147
12	24.37	8.796	4.347
13	6.609	5.752	1.433
14	4.062	1.099	1.085
15	29.59	3.258	1.843

Source: FlintWaterStudy.org (2015) "Lead Results from Tap Water Sampling in Flint, Michigan, during the Flint Water Crisis"

Figure

of 0

Answer 1

Lets assume the resident drinks 3.5 L of water per day

Then in 1 yr = 365 days, total water consumed by her = 3.5 * 365 = 1,277.5‬ L

Then mass of water consumed in 1 yr = density * volume

= 1.0 g/mL * 1,277.5‬ L

= 1Kg / L * 1,277.5‬ L

= 1,277.5 Kg

Now in the first draw data set, the highest amount of lead sample (7) = 40.63 ppb

We know ppb = mass of solute in gram / mass of solvent in 10⁹ gram

i.e mass of solute (lead) present per 10⁹ gram of solvent (water)

Here 40.63 ppb of lead means 40.63 g of lead is present in 10⁹ gram water

Or we can say, in 10⁹ gram = 10⁶ Kg of water, we have lead = 40.63 g

Then in 1,277.5 Kg of water, we have lead = 40.63 g/ 10⁶ Kg * 1,277.5 Kg

= 0.0519 g

= 5.2 * 10^-2 g

i.e mass of lead that the resident would consume over the course of one year = 5.2 * 10^-2 g