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