Question

Question #c4d6f

Answer 1

`1.3 * 10^(18)`

Explanation:

The thing to remember about a concentration expressed in parts per million is that it tells you the number of parts of solute present for every

`10^6 = 1,000,000`

parts of solution/mixture. In your case, you can say that a part of tuna is equal to `"1 oz"`. This means that a concentration of `"1.4 ppm"` will be equal to

`"1.4 ppm Hg" = "1.4 oz Hg"/(10^6color(white)(.)"oz tuna")`

Notice that because the mass of mercury is so small compared to the mass of tuna, you can assume that the mass of the tuna + mercury is equal to the mass of the tuna.

So in order to have a concentration of `"1.4 ppm"` of mercury in tuna, you need to have `"1.4 oz"` of mercury for every `1,000,000` `"oz"` of tuna.

You can thus say that your `"11-oz"` sample of tuna will contain

`11 color(red)(cancel(color(black)("oz tuna"))) * overbrace("1.4 oz Hg"/(10^6color(red)(cancel(color(black)("oz tuna")))))^(color(blue)("= 1.4 ppm Hg")) = 1.54 * 10^(-5)color(white)(.)"oz Hg"`

Now, to find the number of atoms of mercury present in the steak, use the fact that

`"1 oz " ~~ " 28.35 g"`

to convert the mass of mercury to grams.

`1.54 * 10^(-5) color(red)(cancel(color(black)("oz"))) * "28.35 g"/(1color(red)(cancel(color(black)("oz")))) = 4.366 * 10^(-4)color(white)(.)"g"`

Next, use the molar mass of mercury to convert the mass to moles

`4.366 * 10^(-4) color(red)(cancel(color(black)("g"))) * "1 mole Hg"/(200.59 color(red)(cancel(color(black)("g")))) = 2.177 * 10^(-6)color(white)(.)"moles Hg"`

Finally, to convert this to atoms of mercury, use Avogadro's constant

`2.177 * 10^(-6) color(red)(cancel(color(black)("moles Hg"))) * overbrace((6.022 * 10^(23)color(white)(.)"atoms Hg")/(1color(red)(cancel(color(black)("mole Hg")))))^(color(blue)("Avogadro's constant"))`

`= color(darkgreen)(ul(color(black)(1.3 * 10^(18)color(white)(.)"atoms Hg")))`

The answer is rounded to two sig figs.