Question

Question #9ecfa

Answer 1

`"14.5 ppb"`

Explanation:

The first thing to do here is to use the density of the water sample to determine its mass

`12.3 color(red)(cancel(color(black)("cm"^3))) * "1.00 g"/(1color(red)(cancel(color(black)("cm"^3)))) = "12.3 g"`

Now, in order to find the concentration of arsenic in parts per billion, you need to figure out how many grams of arsenic would be present in `10^9` grams of lake water.

Alternatively, you can say that a concentration of `"1 ppb"` is equivalent to `"1 ng"` of arsenic in `"1 g"` of lake water.

In other words, all you need to do now is to use the known composition of the lake water to find how many nanograms of arsenic you have in `"1 g"` of lake water

`1 color(red)(cancel(color(black)("g lake water"))) * overbrace("178.3 ng As"/(12.3color(red)(cancel(color(black)("g lake water")))))^(color(blue)("the given composition")) = "14.5 ng As"`

So, you now know that `"1 g"` of lake water contains `"14.5 ng"` of arsenic, which means that the sample will have a concentration of arsenic equal to

`color(green)(bar(ul(|color(white)(a/a)color(black)("concentration As " = " 14.5 ppb")color(white)(a/a)|)))`

The answer is rounded to three sig figs.