Question

Question #34902

Answer 1

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 `"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 `"15 mL"` sample. More specifically, let's say that the `"15 mL"` sample contains `x` grams of sodium carbonate.

Now, your goal here is to scale up this `"15 mL"` sample to a total volume of `"575 L"` by using the known composition of the sample.

You know that

`"15 mL solution Na"_2"CO"_3 -> x" g Na"_2"CO"_3`

Now, you can go from milliliters to liters by using

`color(darkgreen)(ul(color(black)("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"))`

`= (3.8 * x) * 10^4"g Na"_2"CO"_3`

You plug the value of `x` here and you get the mass of sodium carbonate present in `"575 L"` of solution that contains `x` grams of sodium carbonate in `"15 mL"` of solution.