# A 4.33 g sample of a laboratory solution contains 1.32 g of acid. What is the concentration of the solution as a mass percentage?

Jul 12, 2016

30.5%

#### Explanation:

A solution's percent concentration by mass, $\text{% m/m}$, tells you how many grams of solute, which in your case is an unknown acid, are present in $\text{100 g}$ of solution.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{% m/m" = "grams of solute / 100 g of solution} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

As you know, a solution contains a solute and a solvent, which more often than not is water. This means that a solution's $\text{% m/m}$ concentration tells you how many grams of solute and grams of solvent must be mixed to get $\text{100 g}$ of solution.

In your case, you know that you get $\text{1.32 g}$ of acid in $\text{4.33 g}$ of solution. To find the solution's percent concentration by mass, use this as a conversion factor to figure out how many grams of solute you'd get in $\text{100 g}$ of solution

100 color(red)(cancel(color(black)("g solution"))) * overbrace("1.32 g acid"/(4.33color(red)(cancel(color(black)("g solution")))))^(color(purple)("given to you")) = "30.5 g acid"

Since $\text{100 g}$ of this solution contain $\text{30.5 g}$ of acid, it follows that the percent concentration by mass is

"% m/m" = color(green)(|bar(ul(color(white)(a/a)color(black)("30.5% acid")color(white)(a/a)|)))

The answer is rounded to three sig figs.

Notice that this is equivalent to dividing the mass of solute by the total mass of the solution and multiplying by $100$

"% m/m" = (1.32 color(red)(cancel(color(black)("g acid"))))/(4.33color(red)(cancel(color(black)("g solution")))) * 100 = color(green)(|bar(ul(color(white)(a/a)color(black)("30.5% acid")color(white)(a/a)|)))

As mentioned before, you can use the solution's percent concentration by mass to say that $\text{100 g}$ of solution will contain

$\text{30.5 g acid"" }$ and$\text{ " "100 g " - " 30.5 g" = "69.5 g water}$

In other words, you can make this solution by adding $\text{30.5 g}$ of acid for every $\text{69.5 g}$ of water, which of course gets you $\text{30.5 g}$ of acid per $\text{100 g}$ of solution.