# How many liters of 4.00 M solution can be made using exactly 100.0 grams of lithum bromide?

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the answer is given i just don'y know how to do the work to get the answer?

the answer is given i just don'y know how to do the work to get the answer?

##### 1 Answer

#### Explanation:

The first thing that you need to do here is to convert the mass of lithium bromide to *moles* by using the compound's **molar mass**.

#100.0 color(red)(cancel(color(black)("g"))) * "1 mole LiBr"/(86.845color(red)(cancel(color(black)("g")))) = "1.1515 moles LiBr"#

Now, the **molarity** of the solution is simply a measure of the number of moles of solute, which in your case is lithium bromide, present for every

In order to have a **moles** of lithium bromide **for every**

You know that your sample contains **moles** of lithium bromide, so you can use the molarity of the solution as a *conversion factor* to determine how many liters of this solution can be made.

#1.1515 color(red)(cancel(color(black)("moles LiBr"))) * "1.00 L solution"/(4.00color(red)(cancel(color(black)("moles LiBr")))) = color(darkgreen)(ul(color(black)("0.288 L solution")))#

The answer is rounded to three **sig figs**, the number of sig figs you have for the molarity of the solution.