Question #34902

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.