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).

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
he**r = 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 10 ^{9} gram**

i.e **mass of solute (lead) present per 10 ^{9}
gram of solvent (water)**

Here **40.63 ppb of lead means 40.63 g of lead is present
in 10 ^{9} gram water**

Or we can say, in **10 ^{9} gram = 10^{6}
Kg of water, we have lead = 40.63 g**

Then in **1,277.5 Kg of water, we have lead** =
40.63 g/ 10^{6} 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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