# Question 91ed4

Oct 28, 2016

Here's what I got.

#### Explanation:

Assuming that you're supposed to calculate the solution's percent concentration by mass, $\text{% m/m}$, your starting point here will be to use the density of the solution to calculate its mass.

The solution is said to have a density of ${\text{1.12 g mL}}^{- 1}$, which means that every milliliter of this solution has a mass of $\text{1.12 g}$.

Your sample will thus have a mass of

5.00 * 10^2color(red)(cancel(color(black)("mL"))) * "1.12 g"/(1color(red)(cancel(color(black)("mL")))) = "560 g"

Now, the solution's percent concentration by mass tells you how much solute, which in your case is sodium chloride, $\text{NaCl}$, you get for every $\text{100 g}$ of solution.

Since you know that $\text{560 g}$ of solution contain $\text{21.0 g}$ of sodium chloride, you can say that $\text{100 g}$ of solution will contain

100 color(red)(cancel(color(black)("g solution"))) * "21.0 g NaCl"/(560color(red)(cancel(color(black)("g solution")))) = "3.75 g NaCl"

This means that the solution has a percent concentration by mass equal to

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{% m/m " = " 3.75% NaCl}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.

SIDE NOTE At room temperature, a sodium chloride solution of density equal to ${\text{1.12 g mL}}^{- 1}$ has a percent concentration by mass of about 11%#, so make sure to let me know if this is not the correct interpretation of the problem.

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