An aqueous solution of 4.62 M potassium bromide, KBr, has a density of 1.37 g/mL. What is the percent by mass of KBr in the solution?
1 Answer
Explanation:
Your strategy here will be to

pick a sample volume of this solution and use its density to determine its mass

use the solution's molarity to find how many moles of potassium bromide it contains

use potassium bromide's molar mass to determine how many grams of the compound will contain that many moles
Since you're dealing with molarity, which as you know is defined as moles of solute, in this case potassium bromide,
This sample will have a mass of
#1.00 color(red)(cancel(color(black)("L"))) * (1000 color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.37 g"/(1color(red)(cancel(color(black)("mL")))) = "1370 g"#
Since you know the molarity of this solution, you can say that this
#color(blue)(c = n/V implies n = c * V)#
#n_(KBr) = "4.62 M" * "1.00 L" = "4.62 moles KBr"#
Potassium bromide has a molar mass of
#4.62 color(red)(cancel(color(black)("moles KBr"))) * "119.0 g"/(1color(red)(cancel(color(black)("mole KBr")))) = "549.78 g"#
Finally, a solution's percent concentration by mass is definedas the mass of solute divided by the total mass of the solution, and multiplied by
#color(blue)("% w/w" = "mass of solute"/"mass of solution" xx 100)#
Since this
#"% w/w" = (549.78 color(red)(cancel(color(black)("g"))))/(1370color(red)(cancel(color(black)("g")))) xx 100 = color(green)("40.1%")#
The answer is rounded to three sig figs.