The maximum allowable concentration of lead in drinking water is 9.0 ppb. (b) How many grams of lead are in a swimming pool containing 9.0 ppb lead in 60 m³ of water?

Video transcript

Hi everyone for this problem. It reads the U. S. EPA's regulatory limit for mercury and drinking water is two parts per billion. A 255 liter water sample was acquired from a lake and found to contain 2.3 parts per billion of mercury. How many grams of mercury are there in the water sample? So our goal here is to find the mass of mercury and grams In our water sample. So let's first go ahead and define what is 2. parts per billion of mercury. Okay, so our 2.3 parts per billion of mercury translates to 2.3 g of mercury Per one times 10 to the nine g of water. So we're going to need the mass of water for the liter sample. And to find the mass of water, we're going to need the density of water and the density of water is one g per No leader. Okay. And so let's start off with our 255 liter sample and convert that to grams. Okay, so 255 liters of water because our density is in the unit of millimeter. We need to convert this leader to millimeter first. Okay. And so in one millimeter of water There is 10 to the -3 L of water. Okay. And now we can use the density of water. So in one g there's one mil leader that's the density. Okay, so looking at our units are leaders of water, cancel our milliliters of water cancel. So now we just converted our volume two g. So when we do this calculation, we get 2.55 times To the 5th g of water. So now that we know are massive water, we can figure out our massive mercury by multiplying it by our parts per billion. That was given. Okay. And so our massive water Multiplied by our parts per billion. So we have 2.3 g of mercury Per one times 10 to the 9th g of water. As you can see our units for grams of water cancel. And we're going to be left with mass of mercury. And the question asked us, how many grams of mercury are there in the water sample? So once we do this calculation, we're going to get a final answer of 5.9 times 10 to the negative four g of mercury. And this is going to be our final answer. That's the end of this problem. I hope this was helpful