Question #db9c5

1 Answer
Nov 8, 2017

Answer:

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 #"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.