# Question #42208

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that we use the **percent concentration by mass** to show the number of grams of solute present for every

In this case, you have a solution that is **by mass** silver nitrate,

#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

#13.8 color(red)(cancel(color(black)("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**.