Question

Question #42208

Answer 1

`"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 `"100 g"` of a given solution.

In this case, you have a solution that is `2.23 color(blue)(%)` by mass silver nitrate, `"AgNO"_3`, so you can say that this solution will contain `"2.23 g"` of silver nitrate, the solute, for every `color(blue)("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 `"13.8 g"` of solute.

`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.