# Question c4d6f

Sep 22, 2017

$1.3 \cdot {10}^{18}$

#### 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 $\text{1 oz}$. This means that a concentration of $\text{1.4 ppm}$ 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 $\text{1.4 ppm}$ of mercury in tuna, you need to have $\text{1.4 oz}$ of mercury for every $1 , 000 , 000$ $\text{oz}$ of tuna.

You can thus say that your $\text{11-oz}$ 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

$\text{1 oz " ~~ " 28.35 g}$

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"))#

$= \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{1.3 \cdot {10}^{18} \textcolor{w h i t e}{.} \text{atoms Hg}}}}$

The answer is rounded to two sig figs.