# Question 34902

Jan 16, 2017

Here's what I got.

#### Explanation:

You can't really tell just by the information provided in the problem, but you can set up a conversion factor that can help you figure out how many grams of sodium carbonate you'd get in $\text{575 L}$ of solution.

In order to do that, let's assume that you know how many grams of sodium carbonate you have in the $\text{15 mL}$ sample. More specifically, let's say that the $\text{15 mL}$ sample contains $x$ grams of sodium carbonate.

Now, your goal here is to scale up this $\text{15 mL}$ sample to a total volume of $\text{575 L}$ by using the known composition of the sample.

You know that

${\text{15 mL solution Na"_2"CO"_3 -> x" g Na"_2"CO}}_{3}$

Now, you can go from milliliters to liters by using

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{1 L" = 10^3"mL}}}}$

You can now set up your conversion factors like this

575 color(red)(cancel(color(black)("L solution"))) * (10^3color(red)(cancel(color(black)("mL solution"))))/(1color(red)(cancel(color(black)("L solution")))) * overbrace((x" g Na"_2"CO"_3)/(15color(red)(cancel(color(black)("mL solution")))))^(color(blue)("what you know"))#

$= \left(3.8 \cdot x\right) \cdot {10}^{4} {\text{g Na"_2"CO}}_{3}$

You plug the value of $x$ here and you get the mass of sodium carbonate present in $\text{575 L}$ of solution that contains $x$ grams of sodium carbonate in $\text{15 mL}$ of solution.