# Question #c4d6f

##### 1 Answer

#### 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

#"1.4 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 **for every**

You can thus say that your

#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**.