# 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?

##### 1 Answer

#### Answer:

#### Explanation:

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 **concentration** of

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 **for every**

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

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