# Question 50ff0

May 4, 2016

$\text{8.3 g}$

#### Explanation:

You can use Parts Per Million, or ppm, to express the concentration of solutions that contain very, very small amounts, often called trace amounts, of solute.

More specifically, a concentration of $\text{1 ppm}$ is equivalent to one part solute for every ${10}^{6}$ parts of solvent. One way to express ppm concentration is to use the mass of solute in grams and the mass of solvent in grams

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{ppm" = "grams of solute"/"grams of solvent} \times {10}^{6} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This basically tells you that a concentration of $\text{1 ppm}$ will contain $\text{1 g}$ of solute for every ${10}^{6} \text{g}$ of solvent.

Since you can safely assume that the mass of the solvent is equal to that of the solution, you can say that a con concentration of $\text{6 ppm}$ will contain $\text{6 g}$ of solute in ${10}^{6} \text{g}$ of solution.

Now, the problem tells you this solution contains $\text{0.050 mg}$ of solute, which is equivalent to

0.050 color(red)(cancel(color(black)("mg solute"))) * "1 g"/(10^3color(red)(cancel(color(black)("mg solute")))) = 5.0 * 10^(-5)"g"

Now all you have to do is use the $\text{6 ppm}$ concentration as a conversion factor to determine how many grams of solution would contain this many grams of solvent

$5.0 \cdot {10}^{- 5} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g solute"))) * overbrace((10^6"g solution")/(6color(red)(cancel(color(black)("g solute")))))^(color(purple)("= 6 ppm")) = color(green)(|bar(ul(color(white)(a/a)"8.3 g solution} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

I'll leave the answer rounded to two sig figs.

One interesting thing to notice here is that you can write a concentration of $\text{1 ppm}$ as

$\text{1 g solute"/(10^6"g solution") = (1 * color(red)(cancel(color(black)(10^(3))))"mg solute")/(1 * color(red)(cancel(color(black)(10^(6)))) * color(red)(cancel(color(black)(10^(-3))))"kg solution") = "1 mg solute"/"1 kg solution}$

So, if $\text{1 ppm}$ is equivalent to $\text{1 mg}$ of solute in $\text{1 kg}$ of solution, you can say that $\text{6 ppm}$ will be equivalent to $\text{6 mg}$ of solute in $\text{1 kg}$ of solution.

Therefore,

0.050 color(red)(cancel(color(black)("mg solute"))) * overbrace("1 kg solution"/(6color(red)(cancel(color(black)("mg solute")))))^(color(purple)("= 6 ppm")) = "0.0083 kg solution"#

This will once again be

$0.0083 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{kg"))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = color(green)(|bar(ul(color(white)(a/a)"8.3 g solution} \textcolor{w h i t e}{\frac{a}{a}} |}}}$