Question #db9c5

You can't actually answer this question without knowing the density of the solution.

The problem tells you that salt water is `3.5%` salt by mass, which means that for every `"100 g"` of this solution, you get `"3.5 g"` of salt, the solute.

So you know the mass of salt present in `"100 g"` of this solution, but you don't know the mass of `"100 mL"` of the solution.

In order to find the mass of the sample, you need to know its density. Let's say, for example, that a `3.5%` salt by mass solution has a density of `rho` `"g mL"^(-1)`.

This tells you that every `"1 mL"` of this solution has a mass of `rho` `"g"`. In your case, the sample would have a mass of

`100 color(red)(cancel(color(black)("mL solution"))) * overbrace((rho color(white)(.)"g")/(1color(red)(cancel(color(black)("mL solution")))))^(color(blue)("the density of the solution")) = (100 * rho)color(white)(.)"g"`

So if you get `"3.5 g`' of salt for every `"100 g"` of this solution, you can say that your sample will contain

`(color(blue)(cancel(color(black)(100))) * rho) color(red)(cancel(color(black)("g solution"))) * overbrace("3.5 g salt"/(color(blue)(cancel(color(black)(100)))color(red)(cancel(color(black)("g solution")))))^(color(blue)(" = 3.5% m/m salt")) = (3.5 * rho)color(white)(.)"g salt"`

Now, a sodium chloride solution at room temperature that is `3.5%` salt by mass has density of about

`rho = "1.02 g mL"^(-1) ->` see here, at the bottom of the page.

This means that you will have

`"mass of salt" = (3.5 * 1.02)color(white)(.)"g" = "3.6 g salt"`

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the volume of the solution.