Answer
10454 mL
Work Step by Step
It is given that a Pb-contaminated water sample contains 0.0011 % Pb by mass.
So, 0.0011 g of Pb is present in 100 g of water.
For 115 mg of Pb → ?
= (0.115 X 100)/(0.0011)
= 10454 g of water
Density of water = 1 g/mL
10454 g of water = 10454 mL of water
