Question

Using the highest value of lead from the first draw data set, and assuming a resident...

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"

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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

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