What is the weight/weight % of 9.74g of sodium sulfate to a final volume of 165 mL of solution?

1 Answer
Mar 5, 2018

Here's what I got.


The problem here is that you're missing the density of the solution, which implies that you cannot use its volume to determine its mass.

The solution's percent concentration by mass, #"% m/m"#, tells you the number of grams of solute present for every #"100 g"# of the solution.

You know that the solution contains #"9.74 g"# of sodium sulfate, but you don't know its total mass, so you can't determine how many grams of sodium sulfate you'd get for #"100 g"# of this solution.

Now, let's say that this solution has a density of #rho quad "g mL"^(-1)#, which implies that you get #rho quad "g"# for every #"1 mL"# of the solution.

In this case, the total mass of the solution will be

# 165 color(red)(cancel(color(black)("mL solution"))) * (rho quad "g")/(1color(red)(cancel(color(black)("mL solution")))) = (165 * rho) quad "g"#

So now that you know that you have #"9.64 g"# of sodium sulfate in #(165 * rho) quad "g"# of this solution, you can say that #"100 g"# of the solution will contain

#100 color(red)(cancel(color(black)("g solution"))) * ("9.64 g Na"_2"SO"_4)/((165 * rho) color(red)(cancel(color(black)("g solution")))) = (5.84 * rho) quad "g Na"_2"SO"_4#

This means that the solution's percent concentration by mass is equal to

#"% m/m" = (5.84 * rho)% quad "Na"_2"SO"_4#

So if you find the density of the solution, you can plug in its value and get the solution's percent concentration by mass.