Question

Question #d80bb

Answer 1

Here's how you can do that.

Explanation:

The idea here is that you need to work with the solution's percent concentration by mass, `"%m/m"`, to figure out how much solute you have in your sample.

So, the solution's percent concentration by mass is a measure of how many grams of solute you get in `"100 g"` of solution. Indirectly, this can help you figure out how many grams of solvent you have in `"100 g"` of solution, since

`color(purple)(bar(ul(|color(white)(a/a)color(black)(m_"solution" = m_"solute" + m_"solvent")color(white)(a/a)|)))`

In your case, a `"17.3% m/m"` solution will contain `"17.3 g"` of solute for every `"100 g"` of solution. This is, of course, equivalent to saying that `"100 g"` of this solution contains

`m_"solvent" = overbrace("100 g")^(color(blue)("mass of solution")) - overbrace("17.3 g")^(color(darkgreen)("mass of solute")) = "82.7 g solvent"`

Now all you have to do is scale up this proportion to figure out how many grams of solution would contain `"355.2 g"` of solvent

`355.2 color(red)(cancel(color(black)("g solvent"))) * "100 g solution"/(82.7 color(red)(cancel(color(black)("g solvent")))) = color(green)(bar(ul(|color(white)(a/a)color(black)("429.5 g solution")color(white)(a/a)|)))`

Therefore, the mass of solute must be

`m_"solute" = "429.5 g" - "355.2 g" = color(green)(bar(ul(|color(white)(a/a)color(black)("74.3 g solute")color(white)(a/a)|)))`