Question

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

Answer 1

`"2.0 M"`

Explanation:

Molarity = molar concentration = how many moles in `"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_"LiBr" = "43.4 g"/"86.8 g/mol" = "0.5moles"`

`V_"solution" = "250 mL" = "0.25L"`

So

`c = n_"LiBr"/ V_"solution" = "0.5 moles"/"0.25 L" = "2.0 M"`

Answer 2

`2 \ "M"`

Explanation:

Molarity is defined by the following equation

`"molarity"="moles of solute"/"liters of solution"`

So, we first need to convert `43.4 \ "g"` of `LiBr` into moles.

`LiBr` has a molar mass of `86.845 \ "g/mol"`.

So, `43.4 \ "g"` of `LiBr` 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 \ "L"=1000 \ "mL"`

So, `250 \ "mL"=0.25 \ "L"`

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

`:."molarity"=(0.5 \ "mol")/(0.25 \ "L")=2 \ "mol/L"`

We can also write it as `2 \ "M"`, since `1 \ "M"=1 \ "mol/L"`.