A pitcher contains 2 L of a solution whose concentration is 25 g/L. How many grams of solvent would be in 1L of the solution?

This is a great example of a problem that provides you with information that you do not need.

Notice that you are told that the solution has a volume of #"2 L"# and a concentration of #"25 g L"^(-1)#.

The only thing that you need in order to figure out how many grams of solute you get in any volume of this solution is its concentration.

You don't need to know how much solution you have in the pitcher to answer this problem.

This said, a concentration of #"25 g L"^(-1)# tells you that you get #"25 g"# of solute for every #"1 L"# of solution. As it turns out, you must calculate how many grams of solute you get in #"1 L"# of solution.

Since the concentration of the solution already tells you how many grams of solute you get per liter of solution, you will have

#1 color(red)(cancel(color(black)("L"))) * overbrace("25 g solute"/(1color(red)(cancel(color(black)("L")))))^color(purple)("given concentration") = "25 g solute"#

However, you must round this off to one sig fig, since that's how many sig figs you have in #"1 L"#

#m_"solute" = color(green)(|bar(ul(color(white)(a/a)"30 g"color(white)(a/a)|)))#