# Question 42208

Nov 5, 2017

$\text{619 g}$

#### Explanation:

The idea here is that we use the percent concentration by mass to show the number of grams of solute present for every $\text{100 g}$ of a given solution.

In this case, you have a solution that is 2.23 color(blue)(%) by mass silver nitrate, ${\text{AgNO}}_{3}$, so you can say that this solution will contain $\text{2.23 g}$ of silver nitrate, the solute, for every $\textcolor{b l u e}{\text{100 g}}$ of the solution.

2.23 color(blue)(%)color(white)(.)"AgNO"_3 " "implies" " "2.23 g AgNO"_3 color(white)(.)"per"color(white)(.)color(blue)("100 g")color(white)(.)"of solution"#

So all you have to do here is to use the percent concentration of the solution as a conversion factor to find the mass of this solution that would contain $\text{13.8 g}$ of solute.

$13.8 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g AgNO"_3))) * "100 g solution"/(2.23 color(red)(cancel(color(black)("g AgNO"_3)))) = color(darkgreen)(ul(color(black)("619 g solution}}}}$

The answer is rounded to three sig figs.