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

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

Apr 17, 2018

$\text{0.288 L}$

#### 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 $\text{1.00 L}$ of the solution.

In order to have a $\text{4.00-M}$ solution of lithium bromide, you need to have $4.00$ moles of lithium bromide for every $\text{1.00 L}$ of this solution.

You know that your sample contains $1.1515$ 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.

$1.1515 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles LiBr"))) * "1.00 L solution"/(4.00color(red)(cancel(color(black)("moles LiBr")))) = color(darkgreen)(ul(color(black)("0.288 L solution}}}}$

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