Question

Question #396cc

Answer 1

Here's how you can do that.

Explanation:

Assuming that your solution is `"30.00% m/m"`, you can say that every `"100 g"` of this solution will contain `"30.00 g"` of solute.

This is, of course, equivalent to saying that every `"100 mg"` of this solution will contain `"30.00 mg"` of solute.

Since solutions are homogeneous mixtures, i.e. they have the same composition throughout, you can use the percent concentration of the solution to figure out the mass of solute present in your sample.

`"37,750.00" color(red)(cancel(color(black)("mg solution"))) * overbrace("30.00 mg solute"/(100color(red)(cancel(color(black)("mg solution")))))^(color(blue)("= 30.00% m/m")) = "11,325 mg solute"`

This implies that the mass of the solvent, which can be calculated using

`"mass of solution = mass of solute + mass of solvent"`

will be equal to

`"mass of solvent" = "37,750.00 mg " - " 11,325 mg"`

`"mass of solvent = 26,425 mg"`

I'll leave both values rounded to five sig figs, but keep in mind that you only have four sig figs for the percent concentration.