# Question 8df2c

Nov 24, 2017

Here's what I got.

#### Explanation:

A solution's mass by volume percent concentration, $\text{% m/v}$, tells you the mass of solute present for every $\text{100 mL}$ of the solution.

In your case, you know that your solution contains $\text{30 g}$ of salt, the solute, in $\text{200 mL}$ of the solution, so all you have to do in order to find the solution's mass by volume percent concentration is to figure out how many grams of salt would be present in $\text{100 mL}$ of this solution.

To do that, use the known composition as a conversion factor.

100 color(red)(cancel(color(black)("mL solution"))) * "30 g NaCl"/(200color(red)(cancel(color(black)("mL solution")))) = "15 g NaCl"

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

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{% m/v = 20% NaCl}}}}$

Keep in mind that you must round the answer to one significant figure because that's how many sig figs you have for your data.

To find the concentration of the solution in grams per liter, use the fact that

$\text{1 L" = 10^3 color(white)(.)"mL}$

Since you already know that $\text{200 mL}$ of this solution contain $\text{30 g}$ of salt, you can say that ${10}^{3}$ $\text{mL}$ will contain

10^3 color(red)(cancel(color(black)("mL solution"))) * "30 g NaCl"/(200color(red)(cancel(color(black)("mL solution")))) = "150 g NaCl"#

Therefore, this solution has a concentration of

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{concentration = 200 g L"^(-1)color(white)(.)"NaCl}}}}$

Once again, the answer must be rounded to one significant figure.