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

Mar 5, 2018

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, $\text{% m/m}$, tells you the number of grams of solute present for every $\text{100 g}$ of the solution.

You know that the solution contains $\text{9.74 g}$ 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 $\text{100 g}$ of this solution.

Now, let's say that this solution has a density of $\rho \quad {\text{g mL}}^{- 1}$, which implies that you get $\rho \quad \text{g}$ for every $\text{1 mL}$ 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 $\text{9.64 g}$ of sodium sulfate in $\left(165 \cdot \rho\right) \quad \text{g}$ of this solution, you can say that $\text{100 g}$ 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

${\text{% m/m" = (5.84 * rho)% quad "Na"_2"SO}}_{4}$

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