Question

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

Answer 1

Here's what I got.

Explanation:

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.