# What is the concentration in molarity of a 250 mL solution made with 43.4 grams of LiBr?

Feb 27, 2018

$\text{2.0 M}$

#### Explanation:

Molarity = molar concentration = how many moles in $\text{1 L}$ of solution.

M_("LiBr") = M_"Li" + M_"Br" = "6.9 g/mol" + "79.9 g/mol" = "86.8 g/mol"

(I've used the rounded molar masses)

${n}_{\text{LiBr" = "43.4 g"/"86.8 g/mol" = "0.5moles}}$

${V}_{\text{solution" = "250 mL" = "0.25L}}$

So

$c = {n}_{\text{LiBr"/ V_"solution" = "0.5 moles"/"0.25 L" = "2.0 M}}$

Feb 27, 2018

$2 \setminus \text{M}$

#### Explanation:

Molarity is defined by the following equation

$\text{molarity"="moles of solute"/"liters of solution}$

So, we first need to convert $43.4 \setminus \text{g}$ of $L i B r$ into moles.

$L i B r$ has a molar mass of $86.845 \setminus \text{g/mol}$.

So, $43.4 \setminus \text{g}$ of $L i B r$ is equivalent to

(43.4color(red)cancelcolor(black)"g")/(86.845color(red)cancelcolor(black)"g""/mol")~~0.5 \ "mol"

We can also convert the amount of milliliters to liters.

$1 \setminus \text{L"=1000 \ "mL}$

So, $250 \setminus \text{mL"=0.25 \ "L}$

Now, we have the desired units, we can plug it into the molarity equation.

$\therefore \text{molarity"=(0.5 \ "mol")/(0.25 \ "L")=2 \ "mol/L}$

We can also write it as $2 \setminus \text{M}$, since $1 \setminus \text{M"=1 \ "mol/L}$.