# Question db9c5

Nov 8, 2017

Here's what I got.

#### Explanation:

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 $\text{100 g}$ of this solution, you get $\text{3.5 g}$ of salt, the solute.

So you know the mass of salt present in $\text{100 g}$ of this solution, but you don't know the mass of $\text{100 mL}$ 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 $\rho$ ${\text{g mL}}^{- 1}$.

This tells you that every $\text{1 mL}$ of this solution has a mass of $\rho$ $\text{g}$. 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 $\text{100 g}$ 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

$\rho = {\text{1.02 g mL}}^{- 1} \to$ see here, at the bottom of the page.

This means that you will have

$\text{mass of salt" = (3.5 * 1.02)color(white)(.)"g" = "3.6 g salt}$

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.